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csv2brain / app / src / main / assets / exam / c_2_04.json
ubt on 24 Jan 2024 146 KB 增加习题列表
{"msg":"第四节 数据库系统","code":200,"data":{"currentIndex":null,"examId":null,"examTime":null,"questionList":[{"id":"796237944254910465","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>确定系统边界应在数据库设计的(41)阶段进行;关系规范化是在数据库设计的(42)阶段进行。","analyze":"需求分析阶段的任务是:对现实世界要处理的对象(组织、部门、企业等)进行详细调查,在了解现行系统的概况,确定新系统功能的过程中,确定系统边界、收集支持系统目标的基础数据及其处理方法。","multi":0,"questionType":1,"answer":"A","chooseItem":["796237945152491521"],"itemList":[{"id":"796237945152491521","questionId":"796237944254910465","content":" 需求分析","answer":1,"chooseValue":"A"},{"id":"796237945173463041","questionId":"796237944254910465","content":" 概念设计","answer":0,"chooseValue":"B"},{"id":"796237945198628865","questionId":"796237944254910465","content":" 逻辑设计","answer":0,"chooseValue":"C"},{"id":"796237945215406081","questionId":"796237944254910465","content":" 物理设计","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237833005191169","title":"使用UML进行关系数据库的(5)时,需要设计出表达持久数据的实体类及其联系,并将它们映射为数据库表和视图等。","analyze":"本题考查UML数据库建模的基本知识。<br>基于UML的关系数据库设计分为4个阶段:①业务用例设计是进行数据库的需求分析,使用用例图等建立业务模型;②逻辑数据模型设计是确定应用系统所需的持久数据,设计出关系数据库中表达持久数据的实体类及其联系,并将它们映射为数据库表和视图等;③物理数据模型设计使用组件图、配置图等设计数据库的物理模型;④物理实 现设计根据物理数据模型建立具体数据库环境下的数据库表、视图等。","multi":0,"questionType":1,"answer":"B","chooseItem":["796237833923743745"],"itemList":[{"id":"796237833911160833","questionId":"796237833005191169","content":" 业务用例设计","answer":0,"chooseValue":"A"},{"id":"796237833923743745","questionId":"796237833005191169","content":" 逻辑数据模型设计","answer":1,"chooseValue":"B"},{"id":"796237833932132353","questionId":"796237833005191169","content":" 物理数据模型设计","answer":0,"chooseValue":"C"},{"id":"796237833944715265","questionId":"796237833005191169","content":" 物理实现设计","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237850302500865","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>商业智能系统主要包括数据预处理、建立数据仓库、数据分析和数据展现4个主要阶段,其中(23)是处理海量数据的基础:数据分析是体现系统智能的关键,一般采用(24)和数据挖掘技术。","analyze":"本题主要考查商业智能系统的基本知识。<br>商业智能系统主要包括数据预处理、建立数据仓库、数据分析和数据展现4个主要阶段。数据预处理是整合企业原始数据的第一步,它包括数据的抽取、转换和加载三个过程;建立数据仓库则是处理海量数据的基础,数据分析是体现系统智能的关键,一般采用联机分析处理(OLAP)和数据挖掘技术。OLAP不仅进行数据汇总/聚集,同时还提供切片、切块、下钻、上卷和旋转等数据分析功能,用户可以方便地对海量数据进行多维分析。数据挖掘的目标则是挖掘数据背后隐藏的知识,通过关联分析、聚类和分类等方法建立分析模型,预测企业未来发展趋势和将要面临的问题;在海量数据和分析手段增多的情况下,数据展现主要保障系统分析结构的可视化。","multi":0,"questionType":1,"answer":"A","chooseItem":["796237851195887617"],"itemList":[{"id":"796237851195887617","questionId":"796237850302500865","content":" 联机分析处理","answer":1,"chooseValue":"A"},{"id":"796237851208470529","questionId":"796237850302500865","content":" 建立数据仓库","answer":0,"chooseValue":"B"},{"id":"796237851225247745","questionId":"796237850302500865","content":" 数据分析","answer":0,"chooseValue":"C"},{"id":"796237851242024961","questionId":"796237850302500865","content":" 数据展现","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796238030825345025","title":"采用数据仓库技术进行数据收集时,有时会遇一些略微不一致但可以纠正的数据,纠正这些数据的过程称为()。","analyze":"本题主要考查数据仓库相关的基础知识。<br>在采用数据仓库技术进行数据收集时,有时会遇到一些略微不一致但可以纠正的数据,这时需要采用数据清洗技术对这些不一致的数据进行处理和纠正。","multi":0,"questionType":1,"answer":"C","chooseItem":["796238031760674817"],"itemList":[{"id":"796238031739703297","questionId":"796238030825345025","content":" 数据转换","answer":0,"chooseValue":"A"},{"id":"796238031748091905","questionId":"796238030825345025","content":" 数据抽取","answer":0,"chooseValue":"B"},{"id":"796238031760674817","questionId":"796238030825345025","content":" 数据清洗","answer":1,"chooseValue":"C"},{"id":"796238031773257729","questionId":"796238030825345025","content":" 数据装载","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796238000928346113","title":"<p><strong>请作答第<span style=\"color: red\">3</span>个空。</strong></p>给定关系模式科室K (科室号,科室名,负责人,科室电话)、医生Y (医生号,医生名,性别,科室号,联系电话,家庭地址)和患者B (病历号,患者名,性别,医保号,联系方式),并且1个科室有多名医生,1名医生属于1个科室;1名医生可以为多个患者诊疗,1个患者也可以找多名医生诊疗。<br>科室与医生之间的“所属”联系类型、医生与患者之间的“诊疗”联系类型分别为 (43);其中(44)。下列查询“肝胆科”医生的医生名、联系电话及家庭住址的关系代数表达式中,查询效率最高的是(45) 。","analyze":"根据题意可知一个科室有多名医生,一名医生属于一个科室,所以科室与医生之间的“所属”联系类型为l:n;又因为一名医生可以为多个病人诊疗,一个病人也可以找多名医生诊疗,所以医生与病人之间的“诊疗”联系类型为n:m。<br>当医生与病人之间的“诊疗”联系类型为n:m时,需要转换为一个独立的关系,并将医生号和病历号作为主键。<br>根据关系代数表达式查询优化的准则1 “提早执行选取运算”,即对于有选择运算的表达式,应优化成尽可能先执行选择运算的等价表达式,以得到较小的中间结果,减少运算量和从外存读块的次数。准则2“合并乘积与其后的选择运算为连接运算”,即在表达式中,当乘积运算后面是选择运算时,应该合并为连接运算,使选择与乘积一道完成,以避免做完乘积后,需再扫描一个大的乘积关系进行选择运算。","multi":0,"questionType":1,"answer":"D","chooseItem":["796238001893036033"],"itemList":[{"id":"796238001855287297","questionId":"796238000928346113","content":" <img alt=\"\" width=\"150\" height=\"23\" src=\"https://image.chaiding.com/ruankao/323ec8082342c8532ff67297f9cc0a0d.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"A"},{"id":"796238001863675905","questionId":"796238000928346113","content":" <img alt=\"\" width=\"170\" height=\"24\" src=\"https://image.chaiding.com/ruankao/19976cb1d690dfd9580ebefc553162ae.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"B"},{"id":"796238001880453121","questionId":"796238000928346113","content":" <img alt=\"\" width=\"170\" height=\"23\" src=\"https://image.chaiding.com/ruankao/578c3c903955640ac6b44ccb79c38508.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"C"},{"id":"796238001893036033","questionId":"796238000928346113","content":" <img alt=\"\" width=\"170\" height=\"17\" src=\"https://image.chaiding.com/ruankao/5b844e9d7f3048729a7454137cf70899.jpg?x-oss-process=style/ruankaodaren\">","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237868388339713","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>在数据库设计的需求分析、概念结构设计、逻辑结构设计和物理结构设计的四个阶段中,基本E-R图是(41):数据库逻辑结构设计阶段的主要工作步骤依次为(42)。","analyze":"本题考査数据库设计方面的基础知识。<br>概念结构设计是与数据模型无关的,而一个数据库系统的实现,是以具体的DBMS为基础的,在概念结构设计完成之后,就要依照选用的DBMS,进行该DBMS支持的数据模型相对应的逻辑结构设计。逻辑结构设计即是在概念结构设计的基础上进行数据模型设计,可以是层次、网状模型和关系模型,由于当前的绝大多数DBMS都是基于关系模型的,E-R方法又是概念结构设计的主要方法,如何在全局E-R图基础上进行关系模型的逻辑结构设计成为这一阶段的主要内容。在进行逻辑结构设计时并不考虑数据在某一DBMS下的具体物理实现,即数据是如何在计算机中存储的。逻辑结构设计阶段的主要工作步骤如下图所示。<br><input height=\"149\" src=\"https://image.chaiding.com/ruankao/9b0014271203d173f489bf8b0230f8e9.jpg?x-oss-process=style/ruankaodaren\" type=\"image\" width=\"300\" longdesc=\"undefined\"><br>从上图可见,逻辑结构设计阶段的主要工作步骤依次为:转换为数据模型一关系规范化一模式优化一设计用户子模式。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237869369806849"],"itemList":[{"id":"796237869306892289","questionId":"796237868388339713","content":" 需求分析阶段形成的文档,并作为概念结构设计阶段的设计依据","answer":0,"chooseValue":"A"},{"id":"796237869336252417","questionId":"796237868388339713","content":" 逻辑结构设计阶段形成的文档,并作为概念结构设计阶段的设计依据","answer":0,"chooseValue":"B"},{"id":"796237869369806849","questionId":"796237868388339713","content":" 概念结构设计阶段形成的文档,并作为逻辑结构设计阶段的设计依据","answer":1,"chooseValue":"C"},{"id":"796237869399166977","questionId":"796237868388339713","content":" 概念结构设计阶段形成的文档,并作为物理设计阶段的设计依据","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796238005995065345","title":"若对关系 R (A, B, C, D)、S (C, D,E)进行π<sub>1,2,3,4,7</sub>(σ<sub>3=5^4=6</sub>(R*S))运算,则该关系代数表达式与(45)是等价的。","analyze":"本题考查关系代数运算方面的基础知识。<br>自然联接<img alt=\"\" width=\"23\" height=\"19\" src=\"https://image.chaiding.com/ruankao/4ae32b1eb392e5f750bf9fb70de3c453.jpg?x-oss-process=style/ruankaodaren\">是一种特殊的等值连接,它要求两个关系中进行比较的分量必须是相同的属性组,并且在结果集中将重复属性列去掉。本试题σ<sub>3=5^4=6</sub>(R*S)的含义是R*S 后,选取R和S关系中R.C = S.C^R.D = S.D的元组,再进行R.A、R.B、R.C、R.D和S.E的投影关系运算。可见,该关系运算表达式与R*S是等价的。","multi":0,"questionType":1,"answer":"A","chooseItem":["796238006892646401"],"itemList":[{"id":"796238006892646401","questionId":"796238005995065345","content":" <img alt=\"\" width=\"41\" height=\"18\" src=\"https://image.chaiding.com/ruankao/a6e1766eb197c82e9b2dbf014201a48f.jpg?x-oss-process=style/ruankaodaren\">","answer":1,"chooseValue":"A"},{"id":"796238006905229313","questionId":"796238005995065345","content":" <img alt=\"\" width=\"145\" height=\"18\" src=\"https://image.chaiding.com/ruankao/918fe4136a3b2f68d25d3b5410f8d370.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"B"},{"id":"796238006917812225","questionId":"796238005995065345","content":" <img alt=\"\" width=\"94\" height=\"20\" src=\"https://image.chaiding.com/ruankao/6dab2e3bc8b6d0e750975fdd2ff21145.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"C"},{"id":"796238006930395137","questionId":"796238005995065345","content":" <img alt=\"\" width=\"157\" height=\"21\" src=\"https://image.chaiding.com/ruankao/8f8d658e8fe27a1b89b8c8fa5494f327.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237997908447233","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>给定关系模式科室K (科室号,科室名,负责人,科室电话)、医生Y (医生号,医生名,性别,科室号,联系电话,家庭地址)和患者B (病历号,患者名,性别,医保号,联系方式),并且1个科室有多名医生,1名医生属于1个科室;1名医生可以为多个患者诊疗,1个患者也可以找多名医生诊疗。<br>科室与医生之间的“所属”联系类型、医生与患者之间的“诊疗”联系类型分别为 (43);其中(44)。下列查询“肝胆科”医生的医生名、联系电话及家庭住址的关系代数表达式中,查询效率最高的是(45) 。","analyze":"根据题意可知一个科室有多名医生,一名医生属于一个科室,所以科室与医生之间的“所属”联系类型为l:n;又因为一名医生可以为多个病人诊疗,一个病人也可以找多名医生诊疗,所以医生与病人之间的“诊疗”联系类型为n:m。<br>当医生与病人之间的“诊疗”联系类型为n:m时,需要转换为一个独立的关系,并将医生号和病历号作为主键。<br>根据关系代数表达式查询优化的准则1 “提早执行选取运算”,即对于有选择运算的表达式,应优化成尽可能先执行选择运算的等价表达式,以得到较小的中间结果,减少运算量和从外存读块的次数。准则2“合并乘积与其后的选择运算为连接运算”,即在表达式中,当乘积运算后面是选择运算时,应该合并为连接运算,使选择与乘积一道完成,以避免做完乘积后,需再扫描一个大的乘积关系进行选择运算。","multi":0,"questionType":1,"answer":"B","chooseItem":["796237998915080193"],"itemList":[{"id":"796237998898302977","questionId":"796237997908447233","content":" “诊疗”联系需要转换为一个独立的关系,并将医生号和患者名作为主键","answer":0,"chooseValue":"A"},{"id":"796237998915080193","questionId":"796237997908447233","content":" “诊疗”联系需要转换为一个独立的关系,并将医生号和病历号作为主键","answer":1,"chooseValue":"B"},{"id":"796237998936051713","questionId":"796237997908447233","content":" “所属”联系需要转换为一个独立的关系,并将医生号和科室名作为主键","answer":0,"chooseValue":"C"},{"id":"796237998957023233","questionId":"796237997908447233","content":" “所属”联系需要转换为一个独立的关系,并将医生号和科室号作为主键","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237912831184897","title":"<p><strong>请作答第<span style=\"color: red\">3</span>个空。</strong></p>某公司的商品(商品号,商品名称,生产商,单价)和仓库(仓库号,地址,电话,商品号,库存量)两个实体之间的关系如表1和表2所示。<br><img alt=\"\" width=\"638\" height=\"418\" src=\"https://image.chaiding.com/ruankao/5e2901d0ea700c4aad0a3b5a14515def.jpg?x-oss-process=style/ruankaodaren\"><br>商品关系的主键是(42):仓库关系的主键是(43);仓库关系(44),为了解决这一问题,需要将仓库关系分解为(45)。","analyze":"仓库关系存在冗余、修改操作的不一致,以及插入异常和删除异常。例如,仓库号为“01”的商品有3种,其地址就要重复3次,故存在冗余。","multi":0,"questionType":1,"answer":"D","chooseItem":["796237913829429249"],"itemList":[{"id":"796237913766514689","questionId":"796237912831184897","content":" 无冗余、无插入异常,但存在删除异常","answer":0,"chooseValue":"A"},{"id":"796237913783291905","questionId":"796237912831184897","content":" 无冗余,但存在插入异常和删除异常","answer":0,"chooseValue":"B"},{"id":"796237913808457729","questionId":"796237912831184897","content":" 存在冗余,但不存在修改操作的不一致","answer":0,"chooseValue":"C"},{"id":"796237913829429249","questionId":"796237912831184897","content":" 存在冗余、修改操作的不一致,以及插入异常和删除异常","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237878588887041","title":"数据仓库中数据()的特点是指数据一旦进入数据仓库后,将被长期保留并定期加载和刷新,可以进行各种查询操作,但很少对数据进行修改和删除操作。","analyze":"本题考查数据仓库基本概念。<br>数据仓库拥有以下四个特点:<br>①面向主题:操作型数据库的数据组织面向事务处理任务,各个业务系统之间各自分离,而数据仓库中的数据是按照一定的主题域进行组织。主题是一个抽象的概念,是指用户使用数据仓库进行决策时所关心的重点方面,一个主题通常与多个操作型信息系统相关。<br>②集成性:面向事务处理的操作型数据库通常与某些特定的应用相关,数据库之间相互独立,并且往往是异构的。而数据仓库中的数据是在对原有分散的数据库数据进行抽取、清理的基础上经过系统加工、汇总和整理得到的,必须消除源数据中的不一致性,以保证数据仓库内的信息是关于整个企业的一致的全局信息。<br>③相对稳定性:操作型数据库中的数据通常需要实时更新,数据根据需要及时发生变化。数据仓库的数据主要供企业决策分析之用,所涉及的数据操作主要是数据查询,-旦某个数据进入数据仓库以后,一般情况下将被长期保留,也就是数据仓库中一般有大量的查询操作,但修改和删除操作很少,通常只需要定期的加载、刷新。<br>④反映历史变化:操作型数据库主要关心当前某一个时间段内的数据,而数据仓库中的数据通常包含历史信息,系统记录了企业从过去某一时点(如开始应用数据仓库的时点(到目前的各个阶段的信息,通过这些信息,可以对企业的发展历程和未来趋势做出定量分析和预测。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237879549382657"],"itemList":[{"id":"796237879507439617","questionId":"796237878588887041","content":" 面向主题","answer":0,"chooseValue":"A"},{"id":"796237879524216833","questionId":"796237878588887041","content":" 集成性","answer":0,"chooseValue":"B"},{"id":"796237879549382657","questionId":"796237878588887041","content":" 相对稳定性","answer":1,"chooseValue":"C"},{"id":"796237879570354177","questionId":"796237878588887041","content":" 反映历史变化","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237939171414017","title":"<p><strong>请作答第<span style=\"color: red\">3</span>个空。</strong></p>某销售公司数据库的零件关系(零件号,零件名称,供应商,供应商所在地,库存量)如下表所示,其中同一种零件可由不同的供应商供应,一个供应商可以供应多种零件。零件关系的主键为(43),该关系存在冗余以及插入异常和删除异常等问题。为了解决这一问题需要将零件关系分解为(44),分解后的关系模式可以达到(45)。<br><img alt=\"\" width=\"415\" height=\"184\" src=\"https://image.chaiding.com/ruankao/33fa0f9e00ba3f5392adb4047c328f9c.jpg?x-oss-process=style/ruankaodaren\">","analyze":"原零件关系存在非主属性对码的部分函数依赖:(零件号,供应商)―供应商所在地,但是供应商―供应商所在地,故原关系模式零件非2NF的。分解后的关系模式零件1、零件2和零件3消除了非主属性对码的部分函数依赖,同时不存在传递依赖,故达到3NF。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237940110938113"],"itemList":[{"id":"796237940081577985","questionId":"796237939171414017","content":" INF","answer":0,"chooseValue":"A"},{"id":"796237940098355201","questionId":"796237939171414017","content":" 2NF","answer":0,"chooseValue":"B"},{"id":"796237940110938113","questionId":"796237939171414017","content":" 3NF","answer":1,"chooseValue":"C"},{"id":"796237940127715329","questionId":"796237939171414017","content":" 4NF","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237936306704385","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某销售公司数据库的零件关系(零件号,零件名称,供应商,供应商所在地,库存量)如下表所示,其中同一种零件可由不同的供应商供应,一个供应商可以供应多种零件。零件关系的主键为(43),该关系存在冗余以及插入异常和删除异常等问题。为了解决这一问题需要将零件关系分解为(44),分解后的关系模式可以达到(45)。<br><img alt=\"\" width=\"415\" height=\"184\" src=\"https://image.chaiding.com/ruankao/e60b09f3dffc178e38de54ace2d2b0e5.jpg?x-oss-process=style/ruankaodaren\">","analyze":"关系P存在冗余以及插入异常和删除异常等问题,为了解决这一问题需要将零件关系分解。用户无法查询某零件由哪些供应商供应,原因是分解有损连接的,且不保持函数依赖。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237937221062657"],"itemList":[{"id":"796237937195896833","questionId":"796237936306704385","content":" 零件1(零件号,零件名称,供应商,供应商所在地,库存量)","answer":0,"chooseValue":"A"},{"id":"796237937208479745","questionId":"796237936306704385","content":" 零件1(零件号,零件名称)、零件2(供应商,供应商所在地,库存量)","answer":0,"chooseValue":"B"},{"id":"796237937221062657","questionId":"796237936306704385","content":" 零件1(零件号,零件名称)、零件2(零件号,供应商,库存量)、零件3(供应商,供应商所在地)","answer":1,"chooseValue":"C"},{"id":"796237937233645569","questionId":"796237936306704385","content":" 零件1(零件号,零件名称)、零件2(零件号,库存量)、零件3(供应商,供应商所在地)、零件4(供应商所在地,库存量)","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237906594254849","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某公司的商品(商品号,商品名称,生产商,单价)和仓库(仓库号,地址,电话,商品号,库存量)两个实体之间的关系如表1和表2所示。<br><img alt=\"\" width=\"638\" height=\"418\" src=\"https://image.chaiding.com/ruankao/15d39f1ce3310bc8cba1d5053974c861.jpg?x-oss-process=style/ruankaodaren\"><br>商品关系的主键是(42):仓库关系的主键是(43);仓库关系(44),为了解决这一问题,需要将仓库关系分解为(45)。","analyze":"本题考查的是应试者对关系模式中主键、外键和模式分解及相关知识的掌握程度。<br>试题(42)考查的是关系模式中主键方面的基础知识。商品关系的主键是商品号。","multi":0,"questionType":1,"answer":"A","chooseItem":["796237907680579585"],"itemList":[{"id":"796237907680579585","questionId":"796237906594254849","content":" 商品号","answer":1,"chooseValue":"A"},{"id":"796237907693162497","questionId":"796237906594254849","content":" 商品名称","answer":0,"chooseValue":"B"},{"id":"796237907705745409","questionId":"796237906594254849","content":" 生产商","answer":0,"chooseValue":"C"},{"id":"796237907718328321","questionId":"796237906594254849","content":" 单价","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237915888832513","title":"<p><strong>请作答第<span style=\"color: red\">4</span>个空。</strong></p>某公司的商品(商品号,商品名称,生产商,单价)和仓库(仓库号,地址,电话,商品号,库存量)两个实体之间的关系如表1和表2所示。<br><img alt=\"\" width=\"638\" height=\"418\" src=\"https://image.chaiding.com/ruankao/3679955178896a8912e0aaaad191e546.jpg?x-oss-process=style/ruankaodaren\"><br>商品关系的主键是(42):仓库关系的主键是(43);仓库关系(44),为了解决这一问题,需要将仓库关系分解为(45)。","analyze":"为了解决仓库关系模式存在的问题需要进行模式分解,其中选项A分解存在的问题是仓库2不属于第三范式,因为存在非主属性对码的部分函数依赖,即仓库号一电话。选项B分解存在的问题是分解有损连接,即分解的新关系模式仓库1和仓库2无法恢复到原关系。<br>分解即保持函数依赖,又无损连接,分解的结果如下:<br><img alt=\"\" width=\"623\" height=\"206\" src=\"https://image.chaiding.com/ruankao/98667ed3f151e8411ca3f2a30ca2b641.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"D","chooseItem":["796237916912242689"],"itemList":[{"id":"796237916845133825","questionId":"796237915888832513","content":" 仓库1 (仓库号,地址)和仓库2 (仓库号,电话,商品号,库存量)","answer":0,"chooseValue":"A"},{"id":"796237916874493953","questionId":"796237915888832513","content":" 仓库1 (仓库号,地址,电话〉和仓库2 (商品号,库存量)","answer":0,"chooseValue":"B"},{"id":"796237916899659777","questionId":"796237915888832513","content":" 仓库1 (仓库号,电话)和仓库2 (仓库号,地址,商品号,库存量)","answer":0,"chooseValue":"C"},{"id":"796237916912242689","questionId":"796237915888832513","content":" 仓库1 (仓库号,地址,电话)和仓库2 (仓库号,商品号,库存量)","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237924646539265","title":"<p><strong>请作答第<span style=\"color: red\">3</span>个空。</strong></p>某公司销售数据库的商品、仓库关系模式及函数依赖集F1、F2如下:<br>商品(商品号,商品名称,生产商,单价),F1={商品号→商品名称,商品号→生产商,商品号→单价)},商品关系的主键是(40)。仓库(仓库号,地址,电话,商品号,库存量),F2={仓库号→(地址,电话),(仓库号,商品号)→库存量}。仓库关系的主键是(41),外键是(42)。<br>仓库关系模式(43),为了解决这一问题,需要将仓库关系分解为(44)","analyze":"从仓库关系的函数依赖集F2可以导出(仓库号,商品号)决定仓库关系的全属性, 所以仓库关系的主键是(仓库号,商品号)。又由于商品号是商品关系的主键,故商品号是仓库关系的外键。","multi":0,"questionType":1,"answer":"D","chooseItem":["796237925594451969"],"itemList":[{"id":"796237925544120321","questionId":"796237924646539265","content":" 仓库号","answer":0,"chooseValue":"A"},{"id":"796237925565091841","questionId":"796237924646539265","content":" 地址","answer":0,"chooseValue":"B"},{"id":"796237925577674753","questionId":"796237924646539265","content":" 电话","answer":0,"chooseValue":"C"},{"id":"796237925594451969","questionId":"796237924646539265","content":" 商品号","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796238027876749313","title":"<p><strong>请作答第<span style=\"color: red\">3</span>个空。</strong></p>设有员工实体Employee (员工号,姓名,性别,年龄,电话,家庭住址,家庭成员, 关系,联系电话)。其中,“家庭住址”包括邮编、省、市、街道信息;“家庭成员,关有多个家庭成员。<br>员工实体Employee的主键为(43);该关系属于(44);为使数据库模式设计更合理,对于员工关系模式Employee(45).","analyze":"如果某员工有5个亲属,那么该员工关系中“员工号,姓名,性别,年龄,电话,家庭住址”将重复出现5次,为了将数据库模式设计得更合理,应该消除冗余,即将家庭成员、关系及联系电话加上员工号设计成为一个独立的模式。","multi":0,"questionType":1,"answer":"D","chooseItem":["796238028816273409"],"itemList":[{"id":"796238028770136065","questionId":"796238027876749313","content":" 只允许记录一个亲属的姓名、与员工的关系以及联系电话","answer":0,"chooseValue":"A"},{"id":"796238028791107585","questionId":"796238027876749313","content":" 可以不作任何处理,因为该关系模式达到了3NF","answer":0,"chooseValue":"B"},{"id":"796238028803690497","questionId":"796238027876749313","content":" 增加多个家庭成员、关系及联系电话字段","answer":0,"chooseValue":"C"},{"id":"796238028816273409","questionId":"796238027876749313","content":" 应该将家庭成员、关系及联系电话加上员工号作为一个独立的模式","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796238011317637121","title":"若要使某用户只能查询表EMP中的部分记录,应采取的策略是(41)。","analyze":"本题考查对数据库访问控制方面的基本概念的掌握程度。<br>具有DBA特权的用户可操作数据库的所有资源。<br>将表EMP的查询权限赋予该用户,即全部记录,而题目只允许某用户查询表EMP中的部分记录。<br>编写査询表EMP的存储过程仍然是查询表EMP的所有记录。<br>构建该部分记录的行级视图,并将该视图的査询权限赋予该用户。","multi":0,"questionType":1,"answer":"D","chooseItem":["796238012454293505"],"itemList":[{"id":"796238012399767553","questionId":"796238011317637121","content":" 将该用户级别设定为DBA","answer":0,"chooseValue":"A"},{"id":"796238012416544769","questionId":"796238011317637121","content":" 将表EMP的查询权限赋予该用户","answer":0,"chooseValue":"B"},{"id":"796238012437516289","questionId":"796238011317637121","content":" 编写查询表EMP的存储过程","answer":0,"chooseValue":"C"},{"id":"796238012454293505","questionId":"796238011317637121","content":" 构建该部分记录的行级视图,并将该视图的查询权限赋予该用户","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237973384351745","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>给定关系模式R(U, F),其中,属性集t={城市,街道,邮政编码},函数依赖集F={(城市,街道)→邮政编码,邮政编码→城市}。关系R(41),且分别有(42)。","analyze":"本题考查关系数据库规范化理论方面的基础知识。<br>试题(41)的正确答案是C。因为根据函数依赖定义,可推出(城市,街道)→U,(邮政编码,街道)→U,所以“城市,街道”和“街道,邮政编码”为候选关键字。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237974302904321"],"itemList":[{"id":"796237974273544193","questionId":"796237973384351745","content":" 只有1个候选关键字“城市,街道”","answer":0,"chooseValue":"A"},{"id":"796237974290321409","questionId":"796237973384351745","content":" 只有1个候选关键字“街道,邮政编码”","answer":0,"chooseValue":"B"},{"id":"796237974302904321","questionId":"796237973384351745","content":" 有2个候选关键字“城市,街道”和“街道,邮政编码”","answer":1,"chooseValue":"C"},{"id":"796237974315487233","questionId":"796237973384351745","content":" 有2个候选关键字“城市,街道”和“城市,邮政编码”","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237982007840769","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>给定关系模式R(U,F),U={A,B,C,D},F={AB→C,CD→B}。关系R(42),且分别有(43)。","analyze":"本题考查关系数据库规范化理论方面的基础知识。<br>根据函数依赖定义,可知ACD→U ,ABD→U,所以ACD和ABD均为候选关键字。<br>根据主属性的定义“包含在任何一个候选码中的属性叫做主属性(Prime attribute), 否则叫做非主属性(Nonprime attribute)”,所以,关系R中的4个属性都是主属性。","multi":0,"questionType":1,"answer":"A","chooseItem":["796237982909616129"],"itemList":[{"id":"796237982909616129","questionId":"796237982007840769","content":" 0个非主属性和4个主属性","answer":1,"chooseValue":"A"},{"id":"796237982930587649","questionId":"796237982007840769","content":" 1个非主属性和3个主属性","answer":0,"chooseValue":"B"},{"id":"796237982955753473","questionId":"796237982007840769","content":" 2个非主属性和2个主属性","answer":0,"chooseValue":"C"},{"id":"796237982976724993","questionId":"796237982007840769","content":" 3个非主属性和1个主属性","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237896192380929","title":"数据库的视图与基本表之间,基本表与存储文件之间分别通过建立(40)之间的映像,保证数据的逻辑独立性和物理独立性。","analyze":"本题考查数据库系统管理方面的基础知识。<br>数据库的三级模式结构中,视图对应外模式、基本表对应模式、存储文件对应内模式。数据库系统在三级模式之间提供了两级映像:模式/内模式映像、外模式/模式映像。正因为这两级映像保证了数据库中的数据具有较高的逻辑独立性和物理独立性。<br>①外模式/模式的映像:存在于外部级和概念级之间,实现了外模式到概念模式之间的相互转换。数据的逻辑独立性是指用户的应用程序与数据库的逻辑结构是相互独立的。数据的逻辑结构发生变化后,用户程序也可以不修改。但是,为了保证应用程序能够正确执行,需要修改外模式/概念模式之间的映像。<br>②模式/内模式的映像:存在于概念级和内部级之间,实现了概念模式到内模式之间的相互转换。数据的物理独立性是指当数据库的内模式发生改变时,数据的逻辑结构 不变。由于应用程序处理的只是数据的逻辑结构,这样物理独立性可以保证,当数据的物理结构改变了,应用程序不用改变。但是,为了保证应用程序能够正确执行,需要修改概念模式/内模式之间的映像。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237897178042369"],"itemList":[{"id":"796237897140293633","questionId":"796237896192380929","content":" 模式到内模式和外模式到内模式","answer":0,"chooseValue":"A"},{"id":"796237897161265153","questionId":"796237896192380929","content":" 外模式到内模式和内模式到模式","answer":0,"chooseValue":"B"},{"id":"796237897178042369","questionId":"796237896192380929","content":" 外模式到模式和模式到内模式","answer":1,"chooseValue":"C"},{"id":"796237897199013889","questionId":"796237896192380929","content":" 内模式到模式和模式到外模式","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237909853229057","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某公司的商品(商品号,商品名称,生产商,单价)和仓库(仓库号,地址,电话,商品号,库存量)两个实体之间的关系如表1和表2所示。<br><img alt=\"\" width=\"638\" height=\"418\" src=\"https://image.chaiding.com/ruankao/c1fec906db4abd4ab1affb7c3d67c574.jpg?x-oss-process=style/ruankaodaren\"><br>商品关系的主键是(42):仓库关系的主键是(43);仓库关系(44),为了解决这一问题,需要将仓库关系分解为(45)。","analyze":"本题考查的是应试者对关系模式中主键、外键和模式分解及相关知识的掌握程度。<br>试题(43)考查的是关系模式中主键方面的基础知识。仓库关系的主键是(仓库号,商品号)。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237910796947457"],"itemList":[{"id":"796237910755004417","questionId":"796237909853229057","content":" 仓库号,地址","answer":0,"chooseValue":"A"},{"id":"796237910775975937","questionId":"796237909853229057","content":" 仓库号,电话","answer":0,"chooseValue":"B"},{"id":"796237910796947457","questionId":"796237909853229057","content":" 仓库号,商品号","answer":1,"chooseValue":"C"},{"id":"796237910817918977","questionId":"796237909853229057","content":" 地址,电话","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237871332741121","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>在数据库设计的需求分析、概念结构设计、逻辑结构设计和物理结构设计的四个阶段中,基本E-R图是(41):数据库逻辑结构设计阶段的主要工作步骤依次为(42)。","analyze":"本题考査数据库设计方面的基础知识。<br>概念结构设计是与数据模型无关的,而一个数据库系统的实现,是以具体的DBMS为基础的,在概念结构设计完成之后,就要依照选用的DBMS,进行该DBMS支持的数据模型相对应的逻辑结构设计。逻辑结构设计即是在概念结构设计的基础上进行数据模型设计,可以是层次、网状模型和关系模型,由于当前的绝大多数DBMS都是基于关系模型的,E-R方法又是概念结构设计的主要方法,如何在全局E-R图基础上进行关系模型的逻辑结构设计成为这一阶段的主要内容。在进行逻辑结构设计时并不考虑数据在某一DBMS下的具体物理实现,即数据是如何在计算机中存储的。逻辑结构设计阶段的主要工作步骤如下图所示。<br><input height=\"149\" src=\"https://image.chaiding.com/ruankao/9fee739b0d122420597d6021317e823d.jpg?x-oss-process=style/ruankaodaren\" type=\"image\" width=\"300\" longdesc=\"undefined\"><br>从上图可见,逻辑结构设计阶段的主要工作步骤依次为:转换为数据模型一关系规范化一模式优化一设计用户子模式。","multi":0,"questionType":1,"answer":"B","chooseItem":["796237872293236737"],"itemList":[{"id":"796237872263876609","questionId":"796237871332741121","content":" 关系规范化一转换为数据模型一模式优化一设计用户模式","answer":0,"chooseValue":"A"},{"id":"796237872293236737","questionId":"796237871332741121","content":" 转换为数据模型一关系规范化一模式优化一设计用户模式","answer":1,"chooseValue":"B"},{"id":"796237872322596865","questionId":"796237871332741121","content":" 模式优化一设计用户模式一关系规范化一转换为数据模型","answer":0,"chooseValue":"C"},{"id":"796237872347762689","questionId":"796237871332741121","content":" 设计用户模式一模式优化一关系规范化一转换为数据模型","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237918845816833","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某公司销售数据库的商品、仓库关系模式及函数依赖集F1、F2如下:<br>商品(商品号,商品名称,生产商,单价),F1={商品号→商品名称,商品号→生产商,商品号→单价)},商品关系的主键是(40)。仓库(仓库号,地址,电话,商品号,库存量),F2={仓库号→(地址,电话),(仓库号,商品号)→库存量}。仓库关系的主键是(41),外键是(42)。<br>仓库关系模式(43),为了解决这一问题,需要将仓库关系分解为(44)","analyze":"本题考查应试者对关系模式中主键、外键和模式分解及相关知识的掌握程度。<br>从商品关系的函数依赖集F1可以导出商品号决定商品关系的全属性,所以商品号是商品关系的主键。","multi":0,"questionType":1,"answer":"A","chooseItem":["796237919739203585"],"itemList":[{"id":"796237919739203585","questionId":"796237918845816833","content":" 商品号","answer":1,"chooseValue":"A"},{"id":"796237919764369409","questionId":"796237918845816833","content":" 商品号,商品名称","answer":0,"chooseValue":"B"},{"id":"796237919781146625","questionId":"796237918845816833","content":" 商品号,生产商","answer":0,"chooseValue":"C"},{"id":"796237919793729537","questionId":"796237918845816833","content":" 商品名称,生产商","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237970502864897","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>给定关系模式R(A<sub>1</sub>,A<sub>2</sub>,A<sub>3</sub>,A<sub>4</sub>),R上的函数依赖集F={A<sub>1</sub>A<sub>3</sub>→A<sub>2</sub>,A<sub>2</sub>→A<sub>3</sub>},则R(42)。若将R分解为p={(A<sub>1</sub>A<sub>2</sub>),(A<sub>1</sub>,A<sub>3</sub>)},那么该分解(43)。","analyze":"本题考查关系数据库规范化理论方面的基础知识。<br>因为A<sub>1</sub>A<sub>3</sub>→A2,A<sub>2</sub>→A<sub>3</sub>,没有出现A<sub>4</sub>,所以候选关键字中肯定包A<sub>4</sub>,属性A<sub>1</sub>A<sub>3</sub>A<sub>4</sub>决定全属性,故为候选关键字。同理A<sub>1</sub>A<sub>2</sub>A<sub>4</sub>也为候选关键字。<br>设U1={A1,A2},U2={A1,A3},那么可得出:U1∩U2→(U1-U2)=A<sub>1</sub>→A<sub>2</sub>,U1∩U2→(U2-U1)=A<sub>1</sub>→A<sub>3</sub>,而A<sub>1</sub>-A<sub>2</sub>,A<sub>1</sub>-A<sub>3</sub>∉F<sup>+</sup>,所以分解ρ是有损连接的。<br>又因为F1=F2=∅, F+≠(F1∪F2)<sup>+</sup>,所以分解不保持函数依赖。","multi":0,"questionType":1,"answer":"D","chooseItem":["796237971442388993"],"itemList":[{"id":"796237971396251649","questionId":"796237970502864897","content":" 是无损联接的","answer":0,"chooseValue":"A"},{"id":"796237971417223169","questionId":"796237970502864897","content":" 是保持函数依赖的","answer":0,"chooseValue":"B"},{"id":"796237971429806081","questionId":"796237970502864897","content":" 既是无损联接又保持函数依赖","answer":0,"chooseValue":"C"},{"id":"796237971442388993","questionId":"796237970502864897","content":" 既是有损联接又不保持函数依赖","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237979092799489","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>给定关系模式R(U,F),U={A,B,C,D},F={AB→C,CD→B}。关系R(42),且分别有(43)。","analyze":"本题考查关系数据库规范化理论方面的基础知识。<br>根据函数依赖定义,可知ACD→U ,ABD→U,所以ACD和ABD均为候选关键字。<br>根据主属性的定义“包含在任何一个候选码中的属性叫做主属性(Prime attribute), 否则叫做非主属性(Nonprime attribute)”,所以,关系R中的4个属性都是主属性。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237980044906497"],"itemList":[{"id":"796237980023934977","questionId":"796237979092799489","content":" 只有1个候选关键字ACB","answer":0,"chooseValue":"A"},{"id":"796237980036517889","questionId":"796237979092799489","content":" 只有1个候选关键字BCD","answer":0,"chooseValue":"B"},{"id":"796237980044906497","questionId":"796237979092799489","content":" 有2个候选关键字ACD和ABD","answer":1,"chooseValue":"C"},{"id":"796237980057489409","questionId":"796237979092799489","content":" 有2个候选关键字ACB和BCD","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237863246123009","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>在数据库系统中,数据库的视图、基本表和存储文件的结构分别与(40)对应;数据的物理独立性和数据的逻辑独立性是分别通过修改(41)来完成的。","analyze":"本题考查对数据库基本概念掌握程度。<br>数据库通常采用三级模式结构,其中,视图对应外模式、基本表对应模式、存储文件对应内模式。<br>数据的独立性是由DBMS的二级映像功能来保证的。数据的独立性包括数据的物理独立性和数据的逻辑独立性。数据的物理独立性是指当数据库的内模式发生改变时,数据的逻辑结构不变。为了保证应用程序能够正确执行,需要通过修改概念模式与内模式之间的映像。数据的逻辑独立性是指用户的应用程序与数据库的逻辑结构是相互独立的。数据的逻辑结构发生变化后,用户程序也可以不修改。但是,为了保证应用程序能够正确执行,需要修改外模式与概念模式之间的映像。","multi":0,"questionType":1,"answer":"A","chooseItem":["796237864185647105"],"itemList":[{"id":"796237864185647105","questionId":"796237863246123009","content":" 模式与内模式之间的映像、外模式与模式之间的映像","answer":1,"chooseValue":"A"},{"id":"796237864215007233","questionId":"796237863246123009","content":" 外模式与内模式之间的映像、外模式与模式之间的映像 ","answer":0,"chooseValue":"B"},{"id":"796237864252755969","questionId":"796237863246123009","content":" 外模式与模式之间的映像、模式与内模式之间的映像","answer":0,"chooseValue":"C"},{"id":"796237864282116097","questionId":"796237863246123009","content":" 外模式与内模式之间的映像、模式与内模式之间的映像","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237830077566977","title":"SQL语言支持关系数据库的三级模式结构图如下所示,图中视图、基本表、存储文件分别对应(40)。<br><img alt=\"\" width=\"488\" height=\"273\" src=\"https://image.chaiding.com/ruankao/f308eb3b7a630b77931f0ec4038857f8.jpg?x-oss-process=style/ruankaodaren\">","analyze":"SQL语言支持关系数据库的三级模式结构,其中:视图对应外模式、基本表对应模式、存储文件对应内模式。","multi":0,"questionType":1,"answer":"B","chooseItem":["796237831038062593"],"itemList":[{"id":"796237831012896769","questionId":"796237830077566977","content":" 模式、内模式、外模式","answer":0,"chooseValue":"A"},{"id":"796237831038062593","questionId":"796237830077566977","content":" 外模式、模式、内模式","answer":1,"chooseValue":"B"},{"id":"796237831063228417","questionId":"796237830077566977","content":" 模式、外模式、内模式","answer":0,"chooseValue":"C"},{"id":"796237831084199937","questionId":"796237830077566977","content":" 外模式、内模式、模式","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237967600406529","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>给定关系模式R(A<sub>1</sub>,A<sub>2</sub>,A<sub>3</sub>,A<sub>4</sub>),R上的函数依赖集F={A<sub>1</sub>A<sub>3</sub>→A<sub>2</sub>,A<sub>2</sub>→A<sub>3</sub>},则R(42)。若将R分解为p={(A<sub>1</sub>A<sub>2</sub>),(A<sub>1</sub>,A<sub>3</sub>)},那么该分解(43)。","analyze":"本题考查关系数据库规范化理论方面的基础知识。<br>因为A<sub>1</sub>A<sub>3</sub>→A<sub>2</sub>,A<sub>2</sub>→A<sub>3</sub>,没有出现A<sub>4,</sub>所以候选关键字中肯定包A<sub>4</sub>,属性A<sub>1</sub>A<sub>3</sub>A<sub>4</sub>决定全属性,故为候选关键字。同理A<sub>1</sub>A<sub>2</sub>A<sub>4</sub>也为候选关键字。<br>设U1={A<sub>1</sub>,A<sub>2</sub>},U2={A<sub>1</sub>,A<sub>3</sub>},那么可得出:U1∩U2→(U1-U2)=A<sub>1</sub>→A<sub>2</sub>,U1∩U2→(U2-U1)=A<sub>1</sub>→A<sub>3</sub>,而A<sub>1</sub>-A<sub>2</sub>,A<sub>1</sub>-A<sub>3</sub>∉F<sup>+</sup>,所以分解ρ是有损连接的。<br>又因为F1=F2=∅, F<sup>+</sup>≠(F1∪F2)<sup>+</sup>,所以分解不保持函数依赖。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237968518959105"],"itemList":[{"id":"796237968497987585","questionId":"796237967600406529","content":" 有一个候选关键字A<sub>1</sub>A<sub>3</sub>","answer":0,"chooseValue":"A"},{"id":"796237968510570497","questionId":"796237967600406529","content":" 有一个候选关键字A<sub>1</sub>A<sub>2</sub>A<sub>3</sub>","answer":0,"chooseValue":"B"},{"id":"796237968518959105","questionId":"796237967600406529","content":" 有两个候选关键字A<sub>1</sub>A<sub>3</sub>A<sub>4</sub>和A<sub>1</sub>A<sub>2</sub>A<sub>4</sub>","answer":1,"chooseValue":"C"},{"id":"796237968535736321","questionId":"796237967600406529","content":" 有三个候选关键字A<sub>1</sub>A<sub>2</sub>、A<sub>1</sub>A<sub>3</sub>和A<sub>1</sub>A<sub>4</sub>","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237976257449985","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>给定关系模式R(U, F),其中,属性集t={城市,街道,邮政编码},函数依赖集F={(城市,街道)→邮政编码,邮政编码→城市}。关系R(41),且分别有(42)。","analyze":"根据主属性的定义,“包含在任何一个候选码中的属性叫做主属性(Prime attribute),否则叫做非主属性(Nonprime attribute)”,所以关系中的3个属性都是主属性,而无非主属性。","multi":0,"questionType":1,"answer":"B","chooseItem":["796237977155031041"],"itemList":[{"id":"796237977142448129","questionId":"796237976257449985","content":" 1个非主属性和2个主属性","answer":0,"chooseValue":"A"},{"id":"796237977155031041","questionId":"796237976257449985","content":" 0个非主属性和3个主属性","answer":1,"chooseValue":"B"},{"id":"796237977167613953","questionId":"796237976257449985","content":" 2个非主属性和1个主属性","answer":0,"chooseValue":"C"},{"id":"796237977180196865","questionId":"796237976257449985","content":" 3个非主属性和0个主属性","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237860343664641","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>在数据库系统中,数据库的视图、基本表和存储文件的结构分别与(40)对应;数据的物理独立性和数据的逻辑独立性是分别通过修改(41)来完成的。","analyze":"本题考查对数据库基本概念掌握程度。<br>数据库通常采用三级模式结构,其中,视图对应外模式、基本表对应模式、存储文件对应内模式。<br>数据的独立性是由DBMS的二级映像功能来保证的。数据的独立性包括数据的物理独立性和数据的逻辑独立性。数据的物理独立性是指当数据库的内模式发生改变时,数据的逻辑结构不变。为了保证应用程序能够正确执行,需要通过修改概念模式与内模式之间的映像。数据的逻辑独立性是指用户的应用程序与数据库的逻辑结构是相互独立的。数据的逻辑结构发生变化后,用户程序也可以不修改。但是,为了保证应用程序能够正确执行,需要修改外模式与概念模式之间的映像。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237861274800129"],"itemList":[{"id":"796237861232857089","questionId":"796237860343664641","content":" 模式、外模式、内模式","answer":0,"chooseValue":"A"},{"id":"796237861258022913","questionId":"796237860343664641","content":" 模式、内模式、外模式","answer":0,"chooseValue":"B"},{"id":"796237861274800129","questionId":"796237860343664641","content":" 外模式、模式、内模式","answer":1,"chooseValue":"C"},{"id":"796237861295771649","questionId":"796237860343664641","content":" 外模式、内模式、模式","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237933408440321","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某销售公司数据库的零件关系(零件号,零件名称,供应商,供应商所在地,库存量)如下表所示,其中同一种零件可由不同的供应商供应,一个供应商可以供应多种零件。零件关系的主键为(43),该关系存在冗余以及插入异常和删除异常等问题。为了解决这一问题需要将零件关系分解为(44),分解后的关系模式可以达到(45)。<br><img alt=\"\" width=\"415\" height=\"184\" src=\"https://image.chaiding.com/ruankao/823a8016b40c4065ff1a95faf4085bdc.jpg?x-oss-process=style/ruankaodaren\">","analyze":"根据题意,零件关系的主键为(零件号,供应商)。","multi":0,"questionType":1,"answer":"B","chooseItem":["796237934326992897"],"itemList":[{"id":"796237934306021377","questionId":"796237933408440321","content":" 零件号,零件名称","answer":0,"chooseValue":"A"},{"id":"796237934326992897","questionId":"796237933408440321","content":" 零件号,供应商","answer":1,"chooseValue":"B"},{"id":"796237934347964417","questionId":"796237933408440321","content":" 零件号,供应商所在地","answer":0,"chooseValue":"C"},{"id":"796237934368935937","questionId":"796237933408440321","content":" 供应商,供应商所在地","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237921731497985","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某公司销售数据库的商品、仓库关系模式及函数依赖集F1、F2如下:<br>商品(商品号,商品名称,生产商,单价),F1={商品号→商品名称,商品号→生产商,商品号→单价)},商品关系的主键是(40)。仓库(仓库号,地址,电话,商品号,库存量),F2={仓库号→(地址,电话),(仓库号,商品号)→库存量}。仓库关系的主键是(41),外键是(42)。<br>仓库关系模式(43),为了解决这一问题,需要将仓库关系分解为(44)","analyze":"从仓库关系的函数依赖集F2可以导出(仓库号,商品号)决定仓库关系的全属性, 所以仓库关系的主键是(仓库号,商品号)。又由于商品号是商品关系的主键,故商品号是仓库关系的外键。","multi":0,"questionType":1,"answer":"B","chooseItem":["796237922666827777"],"itemList":[{"id":"796237922654244865","questionId":"796237921731497985","content":" 仓库号","answer":0,"chooseValue":"A"},{"id":"796237922666827777","questionId":"796237921731497985","content":" 仓库号,商品号","answer":1,"chooseValue":"B"},{"id":"796237922687799297","questionId":"796237921731497985","content":" 仓库号,电话","answer":0,"chooseValue":"C"},{"id":"796237922704576513","questionId":"796237921731497985","content":" 地址,电话","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237930489204737","title":"<p><strong>请作答第<span style=\"color: red\">5</span>个空。</strong></p>某公司销售数据库的商品、仓库关系模式及函数依赖集F1、F2如下:<br>商品(商品号,商品名称,生产商,单价),F1={商品号→商品名称,商品号→生产商,商品号→单价)},商品关系的主键是(40)。仓库(仓库号,地址,电话,商品号,库存量),F2={仓库号→(地址,电话),(仓库号,商品号)→库存量}。仓库关系的主键是(41),外键是(42)。<br>仓库关系模式(43),为了解决这一问题,需要将仓库关系分解为(44)","analyze":"了解决仓库关系模式存在的问题,需要进行模式分解。其中,仓库1 (仓库号,地址)和仓库2 (仓库号,电话,商品号,库存量)分解存在的问题是仓库2不属于第三范式,因为存在非主属性对码的部分函数依赖,即仓库号一电话。选项B分解存在的问题是分解有损连接,即分解的新关系模式仓库1和仓库2无法恢复到原关系。仓库1 (仓库号,电话)和仓库2 (仓库号,地址,商品号,库存量)分解存在的问题与A类同,分析略。 仓库1 (仓库号,地址,电话)和仓库2 (仓库号,商品号,库存量)分解是即保持函数依赖,又无损连接,分解的结果如下:<br><img alt=\"\" width=\"442\" height=\"172\" src=\"https://image.chaiding.com/ruankao/4ea55b2683201ac95a462079a2f5aecc.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"D","chooseItem":["796237931479060481"],"itemList":[{"id":"796237931424534529","questionId":"796237930489204737","content":" 仓库1 (仓库号,地址)和仓库2 (仓库号,电话,商品号,库存量)","answer":0,"chooseValue":"A"},{"id":"796237931441311745","questionId":"796237930489204737","content":" 仓库1 (仓库号,地址,电话)和仓库2 (商品号,库存量)","answer":0,"chooseValue":"B"},{"id":"796237931462283265","questionId":"796237930489204737","content":" 仓库1 (仓库号,电话)和仓库2 (仓库号,地址,商品号,库存量)","answer":0,"chooseValue":"C"},{"id":"796237931479060481","questionId":"796237930489204737","content":" 仓库1 (仓库号,地址,电话)和仓库2 (仓库号,商品号,库存量)","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237857491537921","title":"在数据库系统中,为了保证数据库的(40),通常由DBA使用DBMS提供的授权功能为不同用户授权。","analyze":"本题考查数据库安全控制方面的基础知识。<br>数据库管理系统的安全措施有3个方面:<br>①权限机制:通过权限机制,限定用户对数据的操作权限,把数据的操作限定在具有指定权限的用户范围内,以保证数据的安全。在标准SQL中定义了授权语句GRANT来实现权限管理。<br>②视图机制:通过建立用户视图,用户或应用程序只能通过视图来操作数据,保证了视图之外的数据的安全性。<br>③数据加密:对数据库中的数据进行加密,可以防止数据在存储和传输过程中失密。","multi":0,"questionType":1,"answer":"B","chooseItem":["796237858397507585"],"itemList":[{"id":"796237858384924673","questionId":"796237857491537921","content":" 可靠性","answer":0,"chooseValue":"A"},{"id":"796237858397507585","questionId":"796237857491537921","content":" 安全性","answer":1,"chooseValue":"B"},{"id":"796237858410090497","questionId":"796237857491537921","content":" 一致性","answer":0,"chooseValue":"C"},{"id":"796237858422673409","questionId":"796237857491537921","content":" 完整性","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796238033702637569","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>销售公司数据库中的关系零件为P(Pno,Pname,Sname,City,Qty),Pno表示零件号,Pname表示零件名称,Sname表示供应商,City表示所在地,Qty表示库存量。其函数依赖集F={Pno→Pname,(Pno,Sname)→Qty,Sname→City}。关系P为(44),存在冗余度大、修改操作不一致、插入异常和删除异常的问题。若将P分解为(45),则可以解决这一问题。","analyze":"本题考查对数据库基本概念、数据库设计基础知识。<br>原零件关系P存在非主属性对码的部分函数依赖:(Pno, Sname) ——&gt;Qty,但是Pno ——&gt;Pname、Sname ——&gt;City,因此P∈1NF,而非2NF的。1NF主要存在冗余变大、修改操作的不一致、插入异常和删除异常的问题。<br>分解后的关系模式P1P2和P3消除了非主属性对码的部分函数依赖,同时不存在传递依赖,故达到3NF。","multi":0,"questionType":1,"answer":"A","chooseItem":["796238034633773057"],"itemList":[{"id":"796238034633773057","questionId":"796238033702637569","content":" 1NF","answer":1,"chooseValue":"A"},{"id":"796238034671521793","questionId":"796238033702637569","content":" 2NF","answer":0,"chooseValue":"B"},{"id":"796238034709270529","questionId":"796238033702637569","content":" 3NF","answer":0,"chooseValue":"C"},{"id":"796238034751213569","questionId":"796238033702637569","content":" 4NF","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796238021895671809","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>设有员工实体Employee (员工号,姓名,性别,年龄,电话,家庭住址,家庭成员, 关系,联系电话)。其中,“家庭住址”包括邮编、省、市、街道信息;“家庭成员,关有多个家庭成员。<br>员工实体Employee的主键为(43);该关系属于(44);为使数据库模式设计更合理,对于员工关系模式Employee(45).","analyze":"本题考查关系数据库方面的基础知。<br>一个员工可以有多个家庭成员,故为了唯一区分Employee关系中的每一个元组,其主键为(员工号,家庭成员)。","multi":0,"questionType":1,"answer":"B","chooseItem":["796238022965219329"],"itemList":[{"id":"796238022952636417","questionId":"796238021895671809","content":" 员工号","answer":0,"chooseValue":"A"},{"id":"796238022965219329","questionId":"796238021895671809","content":" 员工号,家庭成员","answer":1,"chooseValue":"B"},{"id":"796238022977802241","questionId":"796238021895671809","content":" 姓名","answer":0,"chooseValue":"C"},{"id":"796238022994579457","questionId":"796238021895671809","content":" 姓名,家庭成员","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237927507054593","title":"<p><strong>请作答第<span style=\"color: red\">4</span>个空。</strong></p>某公司销售数据库的商品、仓库关系模式及函数依赖集F1、F2如下:<br>商品(商品号,商品名称,生产商,单价),F1={商品号→商品名称,商品号→生产商,商品号→单价)},商品关系的主键是(40)。仓库(仓库号,地址,电话,商品号,库存量),F2={仓库号→(地址,电话),(仓库号,商品号)→库存量}。仓库关系的主键是(41),外键是(42)。<br>仓库关系模式(43),为了解决这一问题,需要将仓库关系分解为(44)","analyze":"仓库关系存在冗余、插入异常和删除异常,以及修改操作的不一致。例如,仓库号 为“12”的商品有三种,其地址就要重复三次,如下表所示,故仓库关系存在冗余。<br><img alt=\"\" width=\"345\" height=\"164\" src=\"https://image.chaiding.com/ruankao/8d9c9d005791065b34916ed773b96a50.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"A","chooseItem":["796237928442384385"],"itemList":[{"id":"796237928442384385","questionId":"796237927507054593","content":" 存在冗余、插入异常和删除异常,以及修改操作的不一致","answer":1,"chooseValue":"A"},{"id":"796237928454967297","questionId":"796237927507054593","content":" 不存在冗余,但存在插入异常和删除异常","answer":0,"chooseValue":"B"},{"id":"796237928471744513","questionId":"796237927507054593","content":" 不存在修改操作的不一致,但存在冗余和插入异常","answer":0,"chooseValue":"C"},{"id":"796237928488521729","questionId":"796237927507054593","content":" 不存在冗余、插入异常,但存在删除异常和修改操作的不一致","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237893164093441","title":"数据库的视图与基本表之间,基本表与存储文件之间分别通过建立(39)之间的映像,保证数据的逻辑独立性和物理独立性。","analyze":"本题考查数据库系统基本概念。<br>在数据库系统中有三级模式:外模式、模式和内模式。<br>外模式也称为用户模式或子模式,用于描述用户视图层次上的数据特性;模式用于对数据库中全部数据的逻辑结构和特征进行描述,即模式用于描述概念视图层次上的数据特性,如数据库中的基本表;内模式用于描述内部视图层次上的数据特性,是数据在 数据库内部的表示方式,如存储文件。<br>数据库的视图与基本表之间通过外模式到模式之间的映像实现了外模式到概念模式之间的相互转换,即实现了视图与基本表之间的相互转换,从而保证了数据的逻辑独立性。<br>数据库的基本表与存储文件之间通过模式到内模式之间的映像实现了概念模式到内模式之间的相互转换,即实现了基本表与存储文件之间的相互转换,从而保证了数据的物理独立性。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237894254612481"],"itemList":[{"id":"796237894229446657","questionId":"796237893164093441","content":" 模式到内模式和外模式到内模式","answer":0,"chooseValue":"A"},{"id":"796237894242029569","questionId":"796237893164093441","content":" 外模式到内模式和内模式到模式","answer":0,"chooseValue":"B"},{"id":"796237894254612481","questionId":"796237893164093441","content":" 外模式到模式和模式到内模式","answer":1,"chooseValue":"C"},{"id":"796237894263001089","questionId":"796237893164093441","content":" 内模式到模式和模式到外模式","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237881503928321","title":"数据仓库在收集数据过程中,会遇到一些略微不一致但可以纠正的数据,纠正的过程称为(38)。","analyze":"数据仓库从不同的数据源提取数据,各个数据源会存在数据不一致的问题,对少量的略微不一致数据进行纠正(如对地名中的个别错别字等进行纠正),这一概念称为清洗。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237882439258113"],"itemList":[{"id":"796237882409897985","questionId":"796237881503928321","content":" 数据转换","answer":0,"chooseValue":"A"},{"id":"796237882426675201","questionId":"796237881503928321","content":" 数据抽取","answer":0,"chooseValue":"B"},{"id":"796237882439258113","questionId":"796237881503928321","content":" 数据清洗","answer":1,"chooseValue":"C"},{"id":"796237882447646721","questionId":"796237881503928321","content":" 数据装载","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237843117658113","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>关系R、S如下表所示,R÷(π<sub>A1A2</sub>(σ <sub>1&lt;3</sub>(S)))的结果为(44),R、S的左外联接、右外联接和完全外联接的元组个数分别为(45)。<br><img alt=\"\" width=\"331\" height=\"119\" src=\"https://image.chaiding.com/ruankao/44b14606ecad0d388b5cebedcc3e09cc.jpg?x-oss-process=style/ruankaodaren\">","analyze":"两个关系R和S进行自然连接时,选择两个关系R和S公共属性上相等的元组,去掉重复的属性列构成新关系。在这种情况下,关系R中的某些元组有可能在关系S中不存在公共属性值上相等的元组,造成关系R中这些元组的值在运算时舍弃了;同样关系S中的某些元组也可能舍弃。为此,扩充了关系运算左外联接、右外联接和完全外联接。<br>左外联接是指R与S进行自然连接时,只把A中舍弃的元组放到新关系中。<br>右外联接是指R与S进行自然连接时,只把S中舍弃的元组放到新关系中。<br>完全外联接是指R与S进行自然连接时,把尺和^中舍弃的元组都放到新关系中。<br>试题(45)R与S的左外联接、右外联接和完全外联接的结果如下表所示:<br><img alt=\"\" width=\"441\" height=\"250\" src=\"https://image.chaiding.com/ruankao/55f58ec445bf51ec7c801ab3a529decf.jpg?x-oss-process=style/ruankaodaren\"><br>从运算的结果可以看出R与S的左外联接、右外联接和完全外联接的元组个数分别为4,4,6。","multi":0,"questionType":1,"answer":"D","chooseItem":["796237844057182209"],"itemList":[{"id":"796237844019433473","questionId":"796237843117658113","content":" 2,2,4","answer":0,"chooseValue":"A"},{"id":"796237844032016385","questionId":"796237843117658113","content":" 2,2,6","answer":0,"chooseValue":"B"},{"id":"796237844044599297","questionId":"796237843117658113","content":" 4,4,4","answer":0,"chooseValue":"C"},{"id":"796237844057182209","questionId":"796237843117658113","content":" 4,4,6","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237899120005121","title":"数据挖掘的分析方法可以划分为关联分析、序列模式分析、分类分析和聚类分析四种。如果某方法需要一个示例库(该库中的每个元组都有一个给定的类标识)做训练集时,这种分析方法属于()。","analyze":"本题考查数据挖掘基本概念方面的基础知识。<br>数据挖掘就是应用一系列技术从大型数据库或数据仓库中提取人们感兴趣的信息和知识,这些知识或信息是隐含的,事先未知而潜在有用的,提取的知识表示为概念、规则、规律、模式等形式。也可以说,数据挖掘是一类深层次的数据分析。常见和应用最广泛的数据挖掘方法有:<br>①决策树:决策树方法是利用信息论中的互信息(信息增益(寻找数据库中具有最大信息量的属性字段,建立决策树的一个结点,再根据该属性字段的不同取值构建树的分支;在每个分支子集中重复建立树的下层结点和分支的过程。国际上最早的、也是最有影响的决策树方法是Quiulan研究的ID3方法。<br>②神经网络:神经网络方法是模拟人脑神经元结构,完成类似统计学中的判别、回归、聚类等功能,是一种非线性的模型,主要有三种神经网络模型:前馈式网络、反馈式网络和自组织网络。人工神经网络最大的长处是可以自动地从数据中学习,形成知识,这些知识有些是我们过去未曾发现的,因此它具有较强的创新性。神经网络的知识体现在网络连接的权值上,神经网络的学习主要表现在神经网络权值的逐步计算上。<br>③遗传算法:遗传算法是模拟生物进化过程的算法,它由三个基本过程组成:繁殖(选择(、交叉(重组)、变异(突变)。采用遗传算法可以产生优良的后代,经过若干代的遗传,将得到满足要求的后代,即问题得到解决。<br>④关联规则挖掘算法:关联规则是描述数据之间存在关系的规则,形式为A1A2…An=&gt;B1B2…Bn”。一般分为两个步骤:求出大数据项集;用大数据项集产生关联规则。<br>除了上述的常用方法外,还有粗集方法,模糊集合方法,Bayesian Belief Netords,最邻近算法(k-Nearest Neighbors Method(kNN))等。<br>无论采用哪种技术完成数据挖掘,从功能上可以将数据挖掘的分析方法划分为四种,即关联(Associations)分析、序列模式(Sequential Patterns)分析、分类(Classifiers)分析和聚类(Clustering)分析。<br>①关联分析:目的是为了挖掘出隐藏在数据间的相互关系。若设R={A1,A2,…,AP}为{0,1}域上的属性集,r为R上的一个关系,关于r的关联规则表示为X→B,其中X∈R,B∈R,且X∩B=⌑。关联规则的矩阵形式为:矩阵r中,如果在行X的每一列为1,则行B中各列趋向于为1。在进行关联分析的同时还需要计算两个参数,最小置信度(Confidence)和最小支持度(Support)。前者用以过滤掉可能性过小的规则,后者则用来表示这种规则发生的概率,即可信度。<br>②序列模式分析:目的也是为了挖掘出数据之间的联系,但其侧重点在于分析数据间的前后关系(因果关系(。例如,将序列模式分析运用于商业,经过分析,商家可以根据分析结果发现客户潜在的购物模式,发现顾客在购买一种商品的同时经常购买另一种商品的可能性。在进行序列模式分析时也应计算置信度和支持度。<br>③分类分析:首先为每一个记录赋予一个标记(一组具有不同特征的类别X即按标记分类记录,然后检查这些标定的记录,描述出这些记录的特征。这些描述可能是显式的,如一组规则定义;也可能是隐式的,如一个数学模型或公式。<br>④聚类分析:聚类分析法是分类分析法的逆过程,它的输入集是一组未标定的记录,即输入的记录没有作任何处理。目的是根据一定的规则,合理地划分记录集合,并用显式或隐式的方法描述不同的类别。<br>在实际应用的DM系统中,上述四种分析方法有着不同的适用范围,因此经常被综合运用。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237900076306433"],"itemList":[{"id":"796237900055334913","questionId":"796237899120005121","content":" 关联分析","answer":0,"chooseValue":"A"},{"id":"796237900067917825","questionId":"796237899120005121","content":" 序列模式分析","answer":0,"chooseValue":"B"},{"id":"796237900076306433","questionId":"796237899120005121","content":" 分类分析","answer":1,"chooseValue":"C"},{"id":"796237900084695041","questionId":"796237899120005121","content":" 聚类分析","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796238036760285185","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>销售公司数据库中的关系零件为P(Pno,Pname,Sname,City,Qty),Pno表示零件号,Pname表示零件名称,Sname表示供应商,City表示所在地,Qty表示库存量。其函数依赖集F={Pno→Pname,(Pno,Sname)→Qty,Sname→City}。关系P为(44),存在冗余度大、修改操作不一致、插入异常和删除异常的问题。若将P分解为(45),则可以解决这一问题。","analyze":"本题考查对数据库基本概念、数据库设计基础知识。<br>原零件关系P存在非主属性对码的部分函数依赖:(Pno, Sname) ——&gt;Qty,但是Pno ——&gt;Pname、Sname ——&gt;City,因此P∈1NF,而非2NF的。1NF主要存在冗余变大、修改操作的不一致、插入异常和删除异常的问题。<br>分解后的关系模式P1P2和P3消除了非主属性对码的部分函数依赖,同时不存在传递依赖,故达到3NF。","multi":0,"questionType":1,"answer":"C","chooseItem":["796238037708197889"],"itemList":[{"id":"796238037670449153","questionId":"796238036760285185","content":" P1(Pname,Qty)、P2(Pno,Sname,City)","answer":0,"chooseValue":"A"},{"id":"796238037691420673","questionId":"796238036760285185","content":" P1(Pname,Pname)、P2(Sname,City,Qty)","answer":0,"chooseValue":"B"},{"id":"796238037708197889","questionId":"796238036760285185","content":" P1(Pno,Pname)、P2(Pno,Sname,Qty)、P3(Sname,City)","answer":1,"chooseValue":"C"},{"id":"796238037724975105","questionId":"796238036760285185","content":" P1(Pno,Pname)、P2(Pno,Qty)、P3(Sname,City)、P4(City,Qty)","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237947153174529","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>确定系统边界应在数据库设计的(41)阶段进行;关系规范化是在数据库设计的(42)阶段进行。","analyze":"逻辑设计阶段的任务之一是对关系模式进一步地规范化处理。因为生成的初始关系模式并不能完全符合要求,还会有数据冗余、更新异常存在,这就需要根据规范化理论对关系模式分解之后,消除冗余和更新异常。不过有时根据处理要求,可能还需要增加部分冗余以满足处理要求。逻辑设计阶段的任务就需要作部分关系模式的处理,分解、合并或增加冗余属性,提高存储效率和处理效率。","multi":0,"questionType":1,"answer":"C","chooseItem":["796237948126253057"],"itemList":[{"id":"796237948096892929","questionId":"796237947153174529","content":" 需求分析","answer":0,"chooseValue":"A"},{"id":"796237948109475841","questionId":"796237947153174529","content":" 概念设计","answer":0,"chooseValue":"B"},{"id":"796237948126253057","questionId":"796237947153174529","content":" 逻辑设计","answer":1,"chooseValue":"C"},{"id":"796237948134641665","questionId":"796237947153174529","content":" 物理设计","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237994565586945","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>给定关系模式科室K (科室号,科室名,负责人,科室电话)、医生Y (医生号,医生名,性别,科室号,联系电话,家庭地址)和患者B (病历号,患者名,性别,医保号,联系方式),并且1个科室有多名医生,1名医生属于1个科室;1名医生可以为多个患者诊疗,1个患者也可以找多名医生诊疗。<br>科室与医生之间的“所属”联系类型、医生与患者之间的“诊疗”联系类型分别为 (43);其中(44)。下列查询“肝胆科”医生的医生名、联系电话及家庭住址的关系代数表达式中,查询效率最高的是(45) 。","analyze":"根据题意可知一个科室有多名医生,一名医生属于一个科室,所以科室与医生之间的“所属”联系类型为l:n;又因为一名医生可以为多个病人诊疗,一个病人也可以找多名医生诊疗,所以医生与病人之间的“诊疗”联系类型为n:m。<br>当医生与病人之间的“诊疗”联系类型为n:m时,需要转换为一个独立的关系,并将医生号和病历号作为主键。<br>根据关系代数表达式查询优化的准则1 “提早执行选取运算”,即对于有选择运算的表达式,应优化成尽可能先执行选择运算的等价表达式,以得到较小的中间结果,减少运算量和从外存读块的次数。准则2“合并乘积与其后的选择运算为连接运算”,即在表达式中,当乘积运算后面是选择运算时,应该合并为连接运算,使选择与乘积一道完成,以避免做完乘积后,需再扫描一个大的乘积关系进行选择运算。","multi":0,"questionType":1,"answer":"D","chooseItem":["796237995513499649"],"itemList":[{"id":"796237995454779393","questionId":"796237994565586945","content":" 1:1、n:m","answer":0,"chooseValue":"A"},{"id":"796237995475750913","questionId":"796237994565586945","content":" n:m、1:1","answer":0,"chooseValue":"B"},{"id":"796237995492528129","questionId":"796237994565586945","content":" n:m、1:n","answer":0,"chooseValue":"C"},{"id":"796237995513499649","questionId":"796237994565586945","content":" l:n、n:m","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796237840164868097","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>关系R、S如下表所示,R÷(π<sub>A1A2</sub>(σ <sub>1&lt;3</sub>(S)))的结果为(44),R、S的左外联接、右外联接和完全外联接的元组个数分别为(45)。<br><img alt=\"\" width=\"331\" height=\"119\" src=\"https://image.chaiding.com/ruankao/f5cd0865e8ecc819d0bdbaba39516257.jpg?x-oss-process=style/ruankaodaren\">","analyze":"关系代数的除法运算是同时从关系的水平方向和垂直方向进行运算的。若给定关系R(X,Y)和S(Y,Z),X、Y、Z属性组,R÷S应当满足元组在X上的分量值x的象集Y<sub>x</sub>包含S在Y上投影的集合。记作:<br><img alt=\"\" width=\"176\" height=\"27\" src=\"https://image.chaiding.com/ruankao/23de77c72202bffcacf7588274ca62cb.jpg?x-oss-process=style/ruankaodaren\"><br>其中:Y<sub>x</sub>为x在R象集,x=t<sub>r</sub>[X]。且R÷S的结果集的属性组为X。<br>根据除法定义,X属性为A3,Y属性为(A1,A2),R÷S应当满足元组在X上的分量值x的象集Y<sub>x</sub>包含S在Y上投影的集合,所以结果集的属性为A3。属性A3可以取3个值{3,4,7},其中:3的象集为{(1,2)},4的象集为{(2,1),(3,4)},7的象集为{(4,6)}。<br>根据除法定义,本题关系S为(π<sub>A1A2</sub>(σ <sub>1&lt;3</sub>(S)),在属性组Y(A1A2)上的投影为{(2,1),(3,4)}如下表所示:<br><img alt=\"\" width=\"287\" height=\"65\" src=\"https://image.chaiding.com/ruankao/b0a612ebc6ace6c87dccf88ce9c2fc70.jpg?x-oss-process=style/ruankaodaren\"><br>从以上分析可以看出,只有关系R的属性A3的值为4时,其象集包含了关系S在属性组X即(A1,A2)上的投影,所以R÷S={4}。","multi":0,"questionType":1,"answer":"A","chooseItem":["796237841100197889"],"itemList":[{"id":"796237841100197889","questionId":"796237840164868097","content":" {4}","answer":1,"chooseValue":"A"},{"id":"796237841129558017","questionId":"796237840164868097","content":" {3,4}","answer":0,"chooseValue":"B"},{"id":"796237841158918145","questionId":"796237840164868097","content":" {3,4,7}","answer":0,"chooseValue":"C"},{"id":"796237841184083969","questionId":"796237840164868097","content":" {(1,2),(2,1),(3,4),(4,7)}","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796238024949125121","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>设有员工实体Employee (员工号,姓名,性别,年龄,电话,家庭住址,家庭成员, 关系,联系电话)。其中,“家庭住址”包括邮编、省、市、街道信息;“家庭成员,关有多个家庭成员。<br>员工实体Employee的主键为(43);该关系属于(44);为使数据库模式设计更合理,对于员工关系模式Employee(45).","analyze":"关系模式EmployeeeiNF,原因是员工号一(姓名, 性别,年龄,电话,家庭住址),即非主属性(姓名,性别,年龄,电话,家庭住址)不完全依赖于码“员工号,家庭成员”,故Employee不属于2NF。1NF存在4个问题:冗余度大、引起修改操作的不一致性、插入异常和删除异常。","multi":0,"questionType":1,"answer":"D","chooseItem":["796238025926397953"],"itemList":[{"id":"796238025871872001","questionId":"796238024949125121","content":" 2NF,无冗余,无插入异常和删除异常","answer":0,"chooseValue":"A"},{"id":"796238025888649217","questionId":"796238024949125121","content":" 2NF,无冗余,但存在插入异常和删除异","answer":0,"chooseValue":"B"},{"id":"796238025905426433","questionId":"796238024949125121","content":" 1NF,存在冗余,但不存在修改操作的不一致","answer":0,"chooseValue":"C"},{"id":"796238025926397953","questionId":"796238024949125121","content":" 1NF,存在冗余和修改操作的不一致,以及插入异常和删除异常","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233936807153665","title":"在数据库的安全机制中,通过提供()供第三方开发人员调用进行数据更新,从而保证数据库的关系模式不被第三方所获取。","analyze":"在数据库的安全机制中,通过提供存储过程(D)供第三方开发人员调用进行数据更新,从而保证数据库的关系模式不被第三方所获取。存储过程可以封装SQL语句和业务逻辑,并且只允许输入和输出参数与结果,隐藏了底层表的细节,可以保护数据库的安全性。索引(A)是提高数据库查询效率的一种手段;视图(B)是从一个或多个表中筛选出部分数据的虚拟表;触发器(C)是与数据库表事件相关的程序代码,主要用于保持数据完整性,不能保证数据库的关系模式不被第三方所获取。","multi":0,"questionType":1,"answer":"D","chooseItem":["796233937855729665"],"itemList":[{"id":"796233937776037889","questionId":"796233936807153665","content":" 索引","answer":0,"chooseValue":"A"},{"id":"796233937801203713","questionId":"796233936807153665","content":" 视图","answer":0,"chooseValue":"B"},{"id":"796233937826369537","questionId":"796233936807153665","content":" 触发器","answer":0,"chooseValue":"C"},{"id":"796233937855729665","questionId":"796233936807153665","content":" 储存过程","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234416060911617","title":"数据的物理独立性和逻辑独立性分别是通过修改(45)来完成的。","analyze":"本题考查数据独立性方面的基础知识。<br> 数据的独立性是由DBMS的二级映像功能来保证的。数据的独立性包括数据的物理独立性和数据的逻辑独立性。<br> 数据的物理独立性是指当数据库的内模式发生改变时,数据的逻辑结构不变。由于应用程序处理的只是数据的逻辑结构,这样物理独立性可以保证,当数据的物理结构改变了,应用程序不用改变。但是,为了保证应用程序能够正确执行,需要修改概念模式/内模式之间的映像。<br> 数据的逻辑独立性是指用户的应用程序与数据库的逻辑结构是相互独立的。数据的逻辑结构发生变化后,用户程序也可以不修改。但是,为了保证应用程序能够正确执行,需要修改外模式/概念模式之间的映像。","multi":0,"questionType":1,"answer":"D","chooseItem":["796234417214345217"],"itemList":[{"id":"796234417126264833","questionId":"796234416060911617","content":" 外模式与内模式之间的映像、模式与内模式之间的映像","answer":0,"chooseValue":"A"},{"id":"796234417159819265","questionId":"796234416060911617","content":" 外模式与内模式之间的映像、外模式与模式之间的映像","answer":0,"chooseValue":"B"},{"id":"796234417197568001","questionId":"796234416060911617","content":" 外模式与模式之间的映像、模式与内模式之间的映像","answer":0,"chooseValue":"C"},{"id":"796234417214345217","questionId":"796234416060911617","content":" 模式与内模式之间的映像、外模式与模式之间的映像","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234882144555009","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>数据库的产品很多,尽管它们支持的数据模型不同,使用不同的数据库语言,而且数据的存储结构也各不相同,但体系结构基本上都具有相同的特征,采用“三级模式和两级映像”,如下图所示,图中①、②、③分别表示数据库系统中(40),图中④、⑤、⑥分别表示数据库系统中(41)。<br><img alt=\"\" width=\"539\" height=\"363\" src=\"https://image.chaiding.com/ruankao/8a2c58f790379941e46b7d7cfb06359b.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查数据库体系结构基本方而的基础知识。<br>数据库的产品很多,它们支持不同的数据模型,使用不同的数据库语言,建立在不同的操作系统上,而且数据的存储结构也各不相同,但体系结构基本上都具有相同的特征,采用“三级模式和两级映像”。数据库系统采用三级模式结构,这是数据库管理系统内部的系统结构,如下图所示。<br><img width=\"494\" height=\"350\" alt=\"\" src=\"https://image.chaiding.com/ruankao/60ff0e8d4afb94d9c32513220849fda8.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"D","chooseItem":["796234883285405697"],"itemList":[{"id":"796234883209908225","questionId":"796234882144555009","content":" 物理层、逻辑层、视图层","answer":0,"chooseValue":"A"},{"id":"796234883239268353","questionId":"796234882144555009","content":" 逻辑层、物理层、视图层","answer":0,"chooseValue":"B"},{"id":"796234883260239873","questionId":"796234882144555009","content":" 视图层、物理层、逻辑层","answer":0,"chooseValue":"C"},{"id":"796234883285405697","questionId":"796234882144555009","content":" 视图层、逻辑层、物理层","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235121106636801","title":"数据库概念结构设计阶段的工作步骤依次为( )。","analyze":"本题考查数据库系统基本概念。<br>数据库概念结构设计阶段是在需求分析的基础上,依照需求分析中的信息要求,对用户信息加以分类、聚集和概括,建立信息模型,并依照选定的数据库管理系统软件,转换成为数据的逻辑结构,再依照软硬件环境,最终实现数据的合理存储。<br>概念结构设计工作步骤包括:选择局部应用、逐一设计分E-R图和E-R图合并,如下图所示。 <br><img width=\"437\" height=\"226\" alt=\"\" src=\"https://image.chaiding.com/ruankao/9b6e56e5b8c15ebefa0fee01157148cf.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"C","chooseItem":["796235122062938113"],"itemList":[{"id":"796235122016800769","questionId":"796235121106636801","content":" 设计局部视图→抽象数据→修改重构消除冗余→合并取消冲突","answer":0,"chooseValue":"A"},{"id":"796235122037772289","questionId":"796235121106636801","content":" 设计局部视图→抽象数据→合并取消冲突→修改重构消除冗余","answer":0,"chooseValue":"B"},{"id":"796235122062938113","questionId":"796235121106636801","content":" 抽象数据→设计局部视图→合并取消冲突→修改重构消除冗余","answer":1,"chooseValue":"C"},{"id":"796235122088103937","questionId":"796235121106636801","content":" 抽象数据→设计局部视图→修改重构消除冗余→合并取消冲突","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234645170573313","title":"描述企业应用中的实体及其联系,属于数据库设计的( )阶段。","analyze":"本题考查对数据库应用系统设计中各设计阶段的理解。<br> 需求分析用于调查和整理企业数据需求和应用需求;概念设计用于描述企业应用中的实体及其联系;逻辑设计用于逻辑结构的设计,主要是关系模式的设计、视图设计、规范化等;物理设计实现对数据物理组织的描述,包括存取方式、索引设计、数据文件物理分布等。","multi":0,"questionType":1,"answer":"B","chooseItem":["796234646135263233"],"itemList":[{"id":"796234646105903105","questionId":"796234645170573313","content":" 需求分析","answer":0,"chooseValue":"A"},{"id":"796234646135263233","questionId":"796234645170573313","content":" 概念设计","answer":1,"chooseValue":"B"},{"id":"796234646160429057","questionId":"796234645170573313","content":" 逻辑设计","answer":0,"chooseValue":"C"},{"id":"796234646193983489","questionId":"796234645170573313","content":" 物理设计","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235355941523457","title":"在采用三级模式结构的数据库系统中,如果对数据库中的表Emp创建聚簇索引,那么改变的是数据库的(40)。","analyze":"本题考查数据库系统基本概念。<br>内模式也称存储模式,是数据物理结构和存储方式的描述,是数据在数据库内部的表示方式。定义所有的内部记录类型、索引和文件的组织方式,以及数据控制方面的细节。对表Emp创建聚簇索引,意为索引项的顺序是与表中记录的物理顺序一致的索引组织,所以需要改变的是数据库的内模式。","multi":0,"questionType":1,"answer":"B","chooseItem":["796235357099151361"],"itemList":[{"id":"796235357065596929","questionId":"796235355941523457","content":" 模式","answer":0,"chooseValue":"A"},{"id":"796235357099151361","questionId":"796235355941523457","content":" 内模式","answer":1,"chooseValue":"B"},{"id":"796235357128511489","questionId":"796235355941523457","content":" 外模式","answer":0,"chooseValue":"C"},{"id":"796235357157871617","questionId":"796235355941523457","content":" 用户模式","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234399992532993","title":"在数据库系统中,数据的并发控制是指在多用户共享的系统中,协调并发事务的执行, 保证数据库的(40)不受破坏,避免用户得到不正确的数据。","analyze":"本题考查数据库系统的基本概念。<br> 并发控制(Concurrency Control)是指在多用户共享的系统中,许多用户可能同时对同一数据进行操作。DBMS的并发控制子系统负责协调并发事务的执行,保证数据库的完整性不受破坏,避免用户得到不正确的数据。","multi":0,"questionType":1,"answer":"D","chooseItem":["796234401066274817"],"itemList":[{"id":"796234400961417217","questionId":"796234399992532993","content":" 安全性","answer":0,"chooseValue":"A"},{"id":"796234400994971649","questionId":"796234399992532993","content":" 可靠性","answer":0,"chooseValue":"B"},{"id":"796234401032720385","questionId":"796234399992532993","content":" 兼容性","answer":0,"chooseValue":"C"},{"id":"796234401066274817","questionId":"796234399992532993","content":" 完整性","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234885248339969","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>数据库的产品很多,尽管它们支持的数据模型不同,使用不同的数据库语言,而且数据的存储结构也各不相同,但体系结构基本上都具有相同的特征,采用“三级模式和两级映像”,如下图所示,图中①、②、③分别表示数据库系统中(40),图中④、⑤、⑥分别表示数据库系统中(41)。<br><img alt=\"\" width=\"539\" height=\"363\" src=\"https://image.chaiding.com/ruankao/1dadf5056cb8ffb6124bf317760f3044.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查数据库体系结构基本方而的基础知识。<br>数据库的产品很多,它们支持不同的数据模型,使用不同的数据库语言,建立在不同的操作系统上,而且数据的存储结构也各不相同,但体系结构基本上都具有相同的特征,采用“三级模式和两级映像”。数据库系统采用三级模式结构,这是数据库管理系统内部的系统结构,如下图所示。<br><img width=\"494\" height=\"350\" alt=\"\" src=\"https://image.chaiding.com/ruankao/15d986f86d1418ee437e7ae67999d919.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"B","chooseItem":["796234886288527361"],"itemList":[{"id":"796234886246584321","questionId":"796234885248339969","content":" 外模式/内模式映射、外模式/内模式映射、概念模式/内模式映射","answer":0,"chooseValue":"A"},{"id":"796234886288527361","questionId":"796234885248339969","content":" 外模式/概念模式映射、外模式/概念模式映射、概念模式/内模式映射","answer":1,"chooseValue":"B"},{"id":"796234886334664705","questionId":"796234885248339969","content":" 概念模式/内模式映射、概念模式/内模式映射、外模式/内模式映射","answer":0,"chooseValue":"C"},{"id":"796234886380802049","questionId":"796234885248339969","content":" 外模式/内模式映射、外模式/内模式映射、概念模式/外模式映射","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235359242440705","title":"分布式事务的执行可能会涉及到多个站点上的数据操作,在两阶段提交协议中,当事务T<sub>i</sub>的所有读写操作执行结束后,事务Ti的发起者协调器Ci向所有参与T<sub>i</sub>的执行站点发送&lt;prepare T<sub>i</sub>&gt;的消息,当收到所有执行站点返回的&lt;ready T<sub>i</sub>&gt;消息后,Ci再向所有执行站点发送&lt;commit T<sub>i</sub>&gt;消息。若参与事务T<sub>i</sub>执行的某个站点故障恢复后日志中有&lt;ready T<sub>i</sub>&gt;记录,而没有&lt;commit T<sub>i</sub>&gt;记录,则(41)。","analyze":"本题考查数据挖掘基础知识。<br>在两阶段提交协议(Two Phase Commitment Protocol,2PC)中,把分布式事务的某一个代理指定为协调者(Coordinator),所有其他代理称为参与者(Participant)。这里的代理是指完成各个子事务的进程。只有协调者才拥有提交或撤销事务的决定权,而其他参与者各自负责在其本地数据库中执行写操作,并向协调者提出撤销或提交事务的意向。一般一个站点唯一地对应一个子事务,如果某一参与者与协调者在同一站点,虽然它们不需要使用网络来通信,但仍逻辑地认为它与协调者不在同一站点。<br>2PC把事务的提交过程分为两个阶段:<br>第一阶段是表决阶段,目的是形成一个共同的决定。开始时,协调者在它的日志中写入一条开始提交的记录,再给所有参与者发送“准备提交”消息,并进入等待状态。当参与者收到“准备提交”消息后,它检查是否能提交本地事务。如果能提交,参与者在日志中写入一条就绪记录,并给协调发送“建议提交”消息,然后进入就绪状态:否则,参与者写入撤销记录,并给协调者发送“建议撤销”消息。如果某个站点做出“建议撤销”提议,由于撤销决定具有否决权(即单方面撤销),发出“建议撤销”的站点就可以直接忽略这个事务。协调者收到所有参与者的回答后,它就做出是否提交事务的决定。只要有一个参与者建议撤销,协调者就必须从整体上撤销整个分布式事务,因此它写入一条撤销记录,并给所有参与者发送“全局撤销”消息,然后进入撤销状态;否则,它写入提交记录,给所有的参与者发送“全局提交”消息,然后进入提交状态。<br>第二阶段是执行阶段,目的是实现这个协调者的决定。根据协调者的指令,参与者或者提交事务,或者撤销事务,并给协调者发送确认消息。此时,协调者在日志中写入一条事务结束记录并终止事务。<br>本题中,当事务T<sub>i</sub>完成执行时,事务T<sub>i</sub>的发起者协调器C<sub>i</sub>向所有参与T<sub>i</sub>的执行站点发送〈prepare T<sub>i</sub>&gt;的消息,当收到所有执行站点返回的&lt;ready T<sub>i</sub>&gt;消息后,C<sub>i</sub>再向所有执行站点发送&lt;commit T<sub>i</sub>&gt;消息。若参与事务T<sub>i</sub>执行的某个站点故障恢复后日志中有&lt;ready T<sub>i</sub>&gt;记录,而没有&lt;commit T<sub>i</sub>&gt;记录,则应向协调器询问以决定T<sub>i</sub>的最终结果。","multi":0,"questionType":1,"answer":"D","chooseItem":["796235360521703425"],"itemList":[{"id":"796235360391680001","questionId":"796235359242440705","content":" 事务T<sub>i</sub>已完成提交,该站点无需做任何操作","answer":0,"chooseValue":"A"},{"id":"796235360433623041","questionId":"796235359242440705","content":" 事务T<sub>i</sub>已完成提交,该站点应做REDO操作","answer":0,"chooseValue":"B"},{"id":"796235360479760385","questionId":"796235359242440705","content":" 事务T<sub>i</sub>未完成提交,该站点应做UNDO操作","answer":0,"chooseValue":"C"},{"id":"796235360521703425","questionId":"796235359242440705","content":" 应向协调器询问以决定T<sub>i</sub>的最终结果","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235124042649601","title":"设有员工关系Emp (员工号,姓名,性别,年龄,电话,家庭住址,家庭成员,关系,联系电话)。其中,“家庭成员,关系,联系电话”分别记录了员工亲属的姓名、与员工的关系以及联系电话,且一个员工允许有多个家庭成员。为使数据库模式设计更合理,对于员工关系Emp( )。","analyze":"本题考查应试者对关系数据库方面的基础知识。<br>假设某员i有5个亲属,那么该员工关系中“员工号,姓名,性别,年龄,电话,家庭住址”将重复出现5次,为了将数据库模式设计的更合理,应该消除冗余,即将家庭成员、关系及联系电话加上员工号设计成为一个独立的模式。","multi":0,"questionType":1,"answer":"D","chooseItem":["796235125003145217"],"itemList":[{"id":"796235124940230657","questionId":"796235124042649601","content":" 只允许记录一个亲属的姓名、与员工的关系以及联系电话","answer":0,"chooseValue":"A"},{"id":"796235124961202177","questionId":"796235124042649601","content":" 可以不作任何处理,因为该关系模式达到了3NF","answer":0,"chooseValue":"B"},{"id":"796235124982173697","questionId":"796235124042649601","content":" 增加多个家庭成员、关系及联系电话字段","answer":0,"chooseValue":"C"},{"id":"796235125003145217","questionId":"796235124042649601","content":" 应该将家庭成员、关系及联系电话加上员工号设计成一个独立的模式","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234648106586113","title":"某企业信息系统采用分布式数据库系统,该系统中“每节点对本地数据都能独立管理”和“当某一场地故障时,系统可以使用其它场地上的副本而不至于使整个系统瘫痪”分别称为分布式数据库的( )。","analyze":"本题考查对分布式数据库基本概念的理解。<br> 在分布式数据库系统中,共享性是指数据存储在不同的结点数据共享;自治性指每结点对本地数据都能独立管理;可用性是指当某一场地故障时,系统可以使用其他场地上的复本而不至于使整个系统瘫痪;分布性是指数据在不同场地上的存储。","multi":0,"questionType":1,"answer":"C","chooseItem":["796234649109024769"],"itemList":[{"id":"796234649058693121","questionId":"796234648106586113","content":" 共享性和分布性","answer":0,"chooseValue":"A"},{"id":"796234649083858945","questionId":"796234648106586113","content":" 自治性和分布性","answer":0,"chooseValue":"B"},{"id":"796234649109024769","questionId":"796234648106586113","content":" 自治性和可用性","answer":1,"chooseValue":"C"},{"id":"796234649134190593","questionId":"796234648106586113","content":" 分布性和可用性","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235126940913665","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>给定关系模式R&lt;U ,F&gt;,U={A,B,C,D ,E} ,F={B→A ,D→A ,A→E ,AC→B },则R的候选关键字为(42),分解ρ={R<sub>1</sub>(ABCE),R<sub>2</sub>(CD)} (43)。","analyze":"本题考查关系数据库中候选关键字和关系模式的分解问题。<br>根据求属性闭包的算法,可以求得<img alt=\"\" width=\"81\" height=\"21\" src=\"https://image.chaiding.com/ruankao/d92b83541dc8f5040b8ffc56afea462b.jpg?x-oss-process=style/ruankaodaren\">而在CD中不存在一个真子集能决定全属性,故CD为R的候选码。<br>在关系数据库基础理论的相关定义可知,关系模式R&lt;U,F&gt;的一个分解ρ={R<sub>1</sub>(U<sub>1</sub>,F<sub>1</sub>),R<sub>2</sub>(U<sub>2</sub>,F<sub>2</sub>)} ,具有无损连接的充分必要的条件是: <img alt=\"\" width=\"426\" height=\"27\" src=\"https://image.chaiding.com/ruankao/6db34132ac1d2a602135fe1eefc07e37.jpg?x-oss-process=style/ruankaodaren\"><br>根据题意可知:<br><img alt=\"\" width=\"393\" height=\"58\" src=\"https://image.chaiding.com/ruankao/2b2313e392bed270adf2d51b768e2839.jpg?x-oss-process=style/ruankaodaren\"><br>分解ρ={R<sub>1</sub>(ABCE),R<sub>2</sub>(CD)}不满足条件,故不具有无损连接性。<br>又因为<img alt=\"\" width=\"411\" height=\"28\" src=\"https://image.chaiding.com/ruankao/424d283632a003c978d1281a5fca9006.jpg?x-oss-process=style/ruankaodaren\">故分解不保持函数依赖。","multi":0,"questionType":1,"answer":"A","chooseItem":["796235127855271937"],"itemList":[{"id":"796235127855271937","questionId":"796235126940913665","content":" CD","answer":1,"chooseValue":"A"},{"id":"796235127888826369","questionId":"796235126940913665","content":" ABD","answer":0,"chooseValue":"B"},{"id":"796235127918186497","questionId":"796235126940913665","content":" ACD","answer":0,"chooseValue":"C"},{"id":"796235127951740929","questionId":"796235126940913665","content":" ADE","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234651726270465","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>给定关系模式R<U,F>,其中:属性集U = {A,B,C,D,E},函数依赖集F={AC→B,B→DE}。关系R(42),且分别有(43)。","analyze":"AC推出B,B又能推出DE,因此AC是唯一的候选键,其中AC分别是主属性,B、D、E为非主属性。","multi":0,"questionType":1,"answer":"A","chooseItem":["796234652900675585"],"itemList":[{"id":"796234652900675585","questionId":"796234651726270465","content":" 只有1个候选关键字AC","answer":1,"chooseValue":"A"},{"id":"796234652921647105","questionId":"796234651726270465","content":" 只有1个候选关键字AB","answer":0,"chooseValue":"B"},{"id":"796234652951007233","questionId":"796234651726270465","content":" 有2个候选关键字AC和BC","answer":0,"chooseValue":"C"},{"id":"796234652976173057","questionId":"796234651726270465","content":" 无正确答案","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234888419233793","title":"典型的事务服务器系统包括多个在共享内存中访问数据的进程,其中( )监控其它进程,一旦进程失败,它将为该失败进程执行恢复动作,并重启该进程。","analyze":"本题考査数据库系统体系结构基础知识。<br>服务器系统可分为事务服务器和数据服务器系统。典型的事务服务器系统包括多个在共享内存中访问数据的进程,主要有如下类型。<br>•服务器进程是接受用户查询(事务),执行查询并返回结果的进程。<br>•锁管理器进程是实现锁管理器的功能,包括锁授予、所释放和死锁检测。<br>•数据库写进程是将修改过的缓冲块输出到磁盘上。<br>•检査点进程将定期执行检查点操作。<br>•日志写进程是将日志记录写入稳定的存储器上。<br>•进程监控进程是监控其他进程,一旦进程失败,它将为该失败进程执行恢复动作,并重启该进程。","multi":0,"questionType":1,"answer":"C","chooseItem":["796234889614610433"],"itemList":[{"id":"796234889560084481","questionId":"796234888419233793","content":" 检查点进程","answer":0,"chooseValue":"A"},{"id":"796234889589444609","questionId":"796234888419233793","content":" 数据库写进程","answer":0,"chooseValue":"B"},{"id":"796234889614610433","questionId":"796234888419233793","content":" 进程监控进程","answer":1,"chooseValue":"C"},{"id":"796234889639776257","questionId":"796234888419233793","content":" 锁管理器进程","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235362501414913","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>假设关系<em>R</em>(<em>A</em><sub>1</sub>,<em>A</em><sub>2</sub>,<em>A</em><sub>3</sub>)上的一个分解为ρ={(<em>A</em><sub>1</sub>,<em>A</em><sub>2</sub>),(<em>A</em><sub>1</sub>,<em>A</em><sub>3</sub>)},下表是关系<em>R</em>上的一个实例,根据实例推出R的函数依赖集<em>F</em>为(42),分解<em>ρ</em>(43)。<br><img alt=\"\" width=\"184\" height=\"147\" src=\"https://image.chaiding.com/ruankao/cda0568be88860329e2e69e358292f45.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查关系数据库规范化理论基础知识。<br>通过对R上的一个关系实例分析可知,选项A是错误的,因为A<sub>1</sub>→A<sub>2</sub>和A<sub>1</sub>→A<sub>3</sub>是不成立,它们不满足函数依赖的定义。同理选项C和选项D也是错误的。根据候选关键字的定义,不难得出本题的候选关键字是A<sub>1</sub>A<sub>2</sub>和A<sub>1</sub>A<sub>3</sub>,可见试题(42)选项B:F= {A<sub>1</sub>A<sub>3</sub>→A<sub>2</sub>,A<sub>1</sub>A<sub>2</sub>→A<sub>3</sub>}成立。<br>题中分解ρ={(A<sub>1</sub>,A<sub>2</sub>),(A<sub>1</sub>,A<sub>3</sub>)},是有损联接的。<br>关系模式R(U,F)的一个分解ρ={R<sub>1</sub>(U<sub>1</sub>,F<sub>1</sub>),R<sub>2</sub>(U<sub>2</sub>,F<sub>2</sub>)},ρ具有无损联接的充分必要的条件是:U<sub>1</sub>∩U<sub>2</sub>&nbsp;→U1-U<sub>2</sub>∈F<sup>+</sup>或U<sub>1</sub>∩U<sub>2</sub> →U<sub>2</sub>-U<sub>1</sub>∈F<sup>+</sup>","multi":0,"questionType":1,"answer":"B","chooseItem":["796235363415773185"],"itemList":[{"id":"796235363403190273","questionId":"796235362501414913","content":" F={<em>A</em><sub>1</sub>→<em>A</em><sub>2</sub>}","answer":0,"chooseValue":"A"},{"id":"796235363415773185","questionId":"796235362501414913","content":" F={<em>A</em><sub>1</sub><em>A</em><sub>3</sub>→<em>A</em><sub>2,</sub><em>A</em><sub>1</sub><em>A</em><sub>2</sub>→<em>A</em><sub>3</sub>}","answer":1,"chooseValue":"B"},{"id":"796235363424161793","questionId":"796235362501414913","content":" F={<em>A</em><sub>1</sub>→<em>A</em><sub>3</sub>}","answer":0,"chooseValue":"C"},{"id":"796235363436744705","questionId":"796235362501414913","content":" F={<em>A</em><sub>1</sub>→<em>A</em><sub>2</sub>,<em>A</em><sub>1</sub>→<em>A</em><sub>3</sub>}","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234891757899777","title":"给定关系模式R&lt;U,F&gt;,其中U为属性集,F是U上的一组函数依赖,那么Armstrong公理系统的增广律是指( )。","analyze":"本题考查关系数据库基础知识。<br>&nbsp;“若X→Y,X→Z,则X→YZ为F所蕴涵”是Armstrong公理系统的合并规则;<br>&nbsp;“若X→Y,WY→Z,则XW→Z为F所蕴涵”是Armstrong公理系统的伪传递律;<br>&nbsp;“若X→Y,Y→Z为F所蕴涵,则X→Z为F所蕴涵”是Armstrong公理系统的传递律;<br>&nbsp;“若X→Y为F所蕴涵,且Z⊆U,则XZ→YZ为F所蕴涵 ”是Armstrong公理系统的増广律。","multi":0,"questionType":1,"answer":"D","chooseItem":["796234892785504257"],"itemList":[{"id":"796234892710006785","questionId":"796234891757899777","content":" 若X→Y,X→Z,则X→YZ为F所蕴涵","answer":0,"chooseValue":"A"},{"id":"796234892739366913","questionId":"796234891757899777","content":" 若X→Y,WY→Z,则XW→Z为F所蕴涵","answer":0,"chooseValue":"B"},{"id":"796234892764532737","questionId":"796234891757899777","content":" 若X→Y,Y→Z为F所蕴涵,则X→Z为F所蕴涵","answer":0,"chooseValue":"C"},{"id":"796234892785504257","questionId":"796234891757899777","content":" 若X→Y为F所蕴涵,且Z⊆U,则XZ→YZ为F所蕴涵","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234654951690241","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>给定关系模式R<U,F>,其中:属性集U = {A,B,C,D,E},函数依赖集F={AC→B,B→DE}。关系R(42),且分别有(43)。","analyze":"AC推出B,B又能推出DE,因此AC是唯一的候选键,其中AC分别是主属性,B、D、E为非主属性。","multi":0,"questionType":1,"answer":"C","chooseItem":["796234655928963073"],"itemList":[{"id":"796234655866048513","questionId":"796234654951690241","content":" 1个非主属性和4个主属性","answer":0,"chooseValue":"A"},{"id":"796234655899602945","questionId":"796234654951690241","content":" 2个非主属性和3个主属性","answer":0,"chooseValue":"B"},{"id":"796234655928963073","questionId":"796234654951690241","content":" 3个非主属性和2个主属性","answer":1,"chooseValue":"C"},{"id":"796234655962517505","questionId":"796234654951690241","content":" 4个非主属性和1个主属性","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235129897897985","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>给定关系模式R&lt;U ,F&gt;,U={A,B,C,D ,E} ,F={B→A ,D→A ,A→E ,AC→B },则R的候选关键字为(42),分解ρ={R<sub>1</sub>(ABCE),R<sub>2</sub>(CD)} (43)。","analyze":"本题考查关系数据库中候选关键字和关系模式的分解问题。<br>根据求属性闭包的算法,可以求得<img width=\"100\" height=\"26\" alt=\"\" src=\"https://image.chaiding.com/ruankao/b8871184a8ed5b794cd2e2e3954d35ed.jpg?x-oss-process=style/ruankaodaren\">而在CD中不存在一个真子集能决定全属性,故CD为R的候选码。<br>在关系数据库基础理论的相关定义可知,关系模式R&lt;U,F&gt;的一个分解ρ={R<sub>1</sub>(U<sub>1</sub>,F<sub>1</sub>),R<sub>2</sub>(U<sub>2</sub>,F<sub>2</sub>)} ,具有无损连接的充分必要的条件是: <br><img width=\"428\" height=\"28\" alt=\"\" src=\"https://image.chaiding.com/ruankao/2de822a5d1d17915c3a460051105d566.jpg?x-oss-process=style/ruankaodaren\"><br>根据题意可知:<br><img width=\"392\" height=\"58\" alt=\"\" src=\"https://image.chaiding.com/ruankao/291ad5bacc2e73f9571c182fa3ca14ba.jpg?x-oss-process=style/ruankaodaren\"><br>分解ρ={R<sub>1</sub>(ABCE),R<sub>2</sub>(CD)}不满足条件,故不具有无损连接性。<br>又因为<img width=\"410\" height=\"28\" alt=\"\" src=\"https://image.chaiding.com/ruankao/4d9e32949dd0c9d75f54972be5b2fd02.jpg?x-oss-process=style/ruankaodaren\">故分解不保持函数依赖。","multi":0,"questionType":1,"answer":"D","chooseItem":["796235130833227777"],"itemList":[{"id":"796235130791284737","questionId":"796235129897897985","content":" 具有无损连接性,且保持函数依赖","answer":0,"chooseValue":"A"},{"id":"796235130803867649","questionId":"796235129897897985","content":" 不具有无损连接性,但保持函数依赖","answer":0,"chooseValue":"B"},{"id":"796235130820644865","questionId":"796235129897897985","content":" 具有无损连接性,但不保持函数依赖","answer":0,"chooseValue":"C"},{"id":"796235130833227777","questionId":"796235129897897985","content":" 不具有无损连接性,也不保持函数依赖","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235366506975233","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>假设关系<em>R</em>(<em>A</em><sub>1</sub>,<em>A</em><sub>2</sub>,<em>A</em><sub>3</sub>)上的一个分解为ρ={(<em>A</em><sub>1</sub>,<em>A</em><sub>2</sub>),(<em>A</em><sub>1</sub>,<em>A</em><sub>3</sub>)},下表是关系<em>R</em>上的一个实例,根据实例推出R的函数依赖集<em>F</em>为(42),分解<em>ρ</em>(43)。<br><img alt=\"\" width=\"184\" height=\"147\" src=\"https://image.chaiding.com/ruankao/8673ee3bc518446546119e5c9effe815.jpg?x-oss-process=style/ruankaodaren\">","analyze":"在试题(43)中U<sub>1</sub>∩U<sub>2</sub>=A<sub>1</sub>,U<sub>1</sub>-U<sub>2</sub>=A<sub>2</sub>,U<sub>2</sub>-U<sub>1</sub>=A<sub>3</sub>,而A<sub>1</sub>→A<sub>2</sub>∉F<sup>+</sup>和A<sub>1</sub>→A<sub>2</sub>∉F<sup>+</sup>,所以ρ={(A<sub>1</sub>,A<sub>2</sub>)(A<sub>1</sub>,A<sub>3</sub>)}是有损联接的。","multi":0,"questionType":1,"answer":"C","chooseItem":["796235367605882881"],"itemList":[{"id":"796235367526191105","questionId":"796235366506975233","content":" 是无损联接的","answer":0,"chooseValue":"A"},{"id":"796235367572328449","questionId":"796235366506975233","content":" 是保持函数依赖的","answer":0,"chooseValue":"B"},{"id":"796235367605882881","questionId":"796235366506975233","content":" 是有损联接的","answer":1,"chooseValue":"C"},{"id":"796235367622660097","questionId":"796235366506975233","content":" 无法确定是否保持函数依赖","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235132863270913","title":"在关系R(A<sub>1</sub>,A<sub>2</sub> ,A<sub>3</sub>) 和S(A<sub>2</sub>,A<sub>3</sub> ,A<sub>4</sub>) 上进行<img alt=\"\" width=\"215\" height=\"31\" src=\"https://image.chaiding.com/ruankao/0fb14b4a79fdf4393aed9e700fbc1015.jpg?x-oss-process=style/ruankaodaren\">关系运算,与该关系表达式等价的是(44)。","analyze":"<img width=\"620\" height=\"264\" alt=\"\" src=\"https://image.chaiding.com/ruankao/4ab4614bc3e59f2f7df3963f2561b2f2.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"D","chooseItem":["796235133953789953"],"itemList":[{"id":"796235133853126657","questionId":"796235132863270913","content":" <img alt=\"\" width=\"169\" height=\"32\" src=\"https://image.chaiding.com/ruankao/8f35e787714845c0181ea6dc9c57dd89.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"A"},{"id":"796235133899264001","questionId":"796235132863270913","content":" <img alt=\"\" width=\"186\" height=\"30\" src=\"https://image.chaiding.com/ruankao/fade2a8be0439df467ac07251e5b4edf.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"B"},{"id":"796235133937012737","questionId":"796235132863270913","content":" <img alt=\"\" width=\"188\" height=\"31\" src=\"https://image.chaiding.com/ruankao/bea55803aad58291acb126111fa1d8be.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"C"},{"id":"796235133953789953","questionId":"796235132863270913","content":" <img alt=\"\" width=\"227\" height=\"30\" src=\"https://image.chaiding.com/ruankao/5f21ea23f65b7d0371b6106306a7af3b.jpg?x-oss-process=style/ruankaodaren\">","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235369711423489","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>给定关系R(A,B,C,D)和关系S(C,D,E),对其进行自然连接运算R⋈S后的属性列为(44)个;与σ<sub>R.B&gt;S.E</sub>(R⋈S)等价的关系代数表达式为(45)。","analyze":"本题考查关系代数运算方面的知识。<br>自然连接是一种特殊的等值连接,它要求两个关系中进行比较的分量必须是相同的属性组,并且在结果集中将重复属性列去掉。对关系R(A,B,C,D)和关系S(C,D,E)来说,进行等值连接后有7个属性列,去掉2个重复属性列C和D后应为5个,即为R.A,R.B,R.C,R.D,S.E„<br>R×S的属性列为R.A,R.B,R.C,R.D,S.C,S.D,S.E,显然R.A为第1属性列,R.B为第2属性列,R.C为第3属性列,R.D为第4属性列,S.C为第5属性列,S.D为第6属性列,S.E为第7属性列。分析表达式σ<sub>R.B>S.E</sub>(R⋈S)如下:<br>σ<sub>R.B>S.E</sub>等价于σ<sub>2>7<br></sub>R⋈S等价于π<sub>1,2,3,4,7</sub>(σ<sub>3=5∧4=6</sub>(R×S))<br>显然σ<sub>R.B>S.E</sub>(R⋈S)等价于π<sub>1,2,3,4,7</sub>(σ<sub>3=5∧4=6</sub>(R×S))","multi":0,"questionType":1,"answer":"B","chooseItem":["796235370818719745"],"itemList":[{"id":"796235370780971009","questionId":"796235369711423489","content":" 4","answer":0,"chooseValue":"A"},{"id":"796235370818719745","questionId":"796235369711423489","content":" 5","answer":1,"chooseValue":"B"},{"id":"796235370852274177","questionId":"796235369711423489","content":" 6","answer":0,"chooseValue":"C"},{"id":"796235370877440001","questionId":"796235369711423489","content":" 7","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234657988366337","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>若要将部门表Demp中name列的修改权限赋予用户Ming,并允许Ming将该限授予他人,实现的SQL语句如下:<br> GRANT(44)ON TABLE Demp TO Ming(45)。","analyze":"本题考查对标准SQL授权语句的掌握。<br> 标准SQL中授权的语句格式如下:<br> <img width=\"548\" height=\"56\" src=\"https://image.chaiding.com/ruankao/493f3e4e421a3a07a1a174231dc69fce.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> 若在授权时指定了WITH GRANT OPTION,那么获得了权限的用户还可以将权限赋给其他用户。不同类型的操作对象有不同的操作权限,常见的操作权限如表所示。<br> 表常见的操作权限<br> <img width=\"552\" height=\"160\" src=\"https://image.chaiding.com/ruankao/a68892d9c30970576ecf4585fbb5d28a.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> 按试题要求,是要将修改属性列name权限给用户Ming,故空(44)应填写UPDATE(name)。故空(45)应填写WITH GRANT OPTION。","multi":0,"questionType":1,"answer":"B","chooseItem":["796234659020165121"],"itemList":[{"id":"796234658990804993","questionId":"796234657988366337","content":" SELECT(name)","answer":0,"chooseValue":"A"},{"id":"796234659020165121","questionId":"796234657988366337","content":" UPDATE(name)","answer":1,"chooseValue":"B"},{"id":"796234659053719553","questionId":"796234657988366337","content":" INSERT(name)","answer":0,"chooseValue":"C"},{"id":"796234659074691073","questionId":"796234657988366337","content":" ALL PRIVILEGES(name)","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234894719078401","title":"某集团公司下属有多个超市,假设公司高管需要从时间、地区和商品种类三个维度来分析某电器商品销售数据,那么应采用( )来完成。","analyze":"本题考査数据仓库基础知识。<br>ETL(Extract-Transfomi-Load)用来描述将数据从来源端经过抽取(extract)、转换(transform)、加载(load)至目的端的过程。ETL是构建数据仓库的重要环,用户从数椐源抽取出所需的数据,经过数据清洗,最终按照预先定义好的数据仓库模型,将数据加载到数据仓库中去。<br>联机事务处理过程(On-Line Transaction Processing, OLTP)也称为面向交易的处理过程,其基本特征是前台接收的用户数据可以立即传送到计算中心进行处理,并在很短的时间内给出处理结果,是对用户操作快速响应的方式之一。<br>数掘挖掘和联机分析处理(On-Line Analytical Processing, OLAP)同为为分析工具,其差别在于OLAP提供给用户一个便利的多维度观点和方法,以有效率地对数据进行复杂的查询动作,其预设查询条件由用户预先设定,而数据挖掘,则能由资讯系统主动发掘资料来源中未曾被察觉的隐藏资讯和透过用户的认知以产生信息。","multi":0,"questionType":1,"answer":"B","chooseItem":["796234895658602497"],"itemList":[{"id":"796234895633436673","questionId":"796234894719078401","content":" 数据挖掘","answer":0,"chooseValue":"A"},{"id":"796234895658602497","questionId":"796234894719078401","content":" OLAP","answer":1,"chooseValue":"B"},{"id":"796234895687962625","questionId":"796234894719078401","content":" OLTP","answer":0,"chooseValue":"C"},{"id":"796234895708934145","questionId":"796234894719078401","content":" ETL","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235136390680577","title":"<img src=\"https://image.chaiding.com/ruankao/2e47d1a777f20d5a9f903f67ea5e5977.jpg?x-oss-process=style/ruankaodaren\"><br>将该关系代数表达式转换为等价的SQL语句如下:<br><img alt=\"\" width=\"384\" height=\"24\" src=\"https://image.chaiding.com/ruankao/7735b7dfad3cf33a62c90cb9330f3acd.jpg?x-oss-process=style/ruankaodaren\">","analyze":"<img width=\"610\" height=\"117\" alt=\"\" src=\"https://image.chaiding.com/ruankao/7cff7a539a1b5c6abe6442db877efcad.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"C","chooseItem":["796235137497976833"],"itemList":[{"id":"796235137468616705","questionId":"796235136390680577","content":" OR S.A<sub>4</sub> = '95' OR R.A<sub>2</sub> = S.A<sub>2</sub> OR R.A<sub>3</sub> = S.A<sub>3</sub>","answer":0,"chooseValue":"A"},{"id":"796235137485393921","questionId":"796235136390680577","content":" AND S.A<sub>4</sub> = '95' OR R.A<sub>2</sub> = S.A<sub>2</sub> AND R.A<sub>3</sub> = S.A<sub>3</sub>","answer":0,"chooseValue":"B"},{"id":"796235137497976833","questionId":"796235136390680577","content":" AND S.A<sub>4</sub> = '95' AND R.A<sub>2</sub> = S.A<sub>2</sub> AND R.A<sub>3</sub> = S.A<sub>3</sub>","answer":1,"chooseValue":"C"},{"id":"796235137518948353","questionId":"796235136390680577","content":" OR S.A<sub>4</sub> = '95' AND R.A<sub>2</sub> = S.A<sub>2</sub> OR R.A<sub>3</sub> = S.A<sub>3</sub>","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235373066866689","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>给定关系R(A,B,C,D)和关系S(C,D,E),对其进行自然连接运算R⋈S后的属性列为(44)个;与σ<sub>R.B&gt;S.E</sub>(R⋈S)等价的关系代数表达式为(45)。","analyze":"本题考查关系代数运算方面的知识。<br>自然连接是一种特殊的等值连接,它要求两个关系中进行比较的分量必须是相同的属性组,并且在结果集中将重复属性列去掉。对关系R(A,B,C,D)和关系S(C,D,E)来说,进行等值连接后有7个属性列,去掉2个重复属性列C和D后应为5个,即为R.A,R.B,R.C,R.D,S.E„<br>R×S的属性列为R.A,R.B,R.C,R.D,S.C,S.D,S.E,显然R.A为第1属性列,R.B为第2属性列,R.C为第3属性列,R.D为第4属性列,S.C为第5属性列,S.D为第6属性列,S.E为第7属性列。分析表达式σ<sub>R.B>S.E</sub>(R⋈S)如下:<br>σ<sub>R.B>S.E</sub>等价于σ<sub>2>7</sub><br>R⋈S等价于π<sub>1,2,3,4,7</sub>(σ<sub>3=5∧4=6</sub>(R×S))<br>显然σ<sub>R.B>S.E</sub>(R⋈S)等价于π<sub>1,2,3,4,7</sub>(σ<sub>2>7Λ3=5∧4=6</sub>(R×S))","multi":0,"questionType":1,"answer":"D","chooseItem":["796235374169968641"],"itemList":[{"id":"796235374081888257","questionId":"796235373066866689","content":" σ<sub>2>7</sub>(R×S)","answer":0,"chooseValue":"A"},{"id":"796235374111248385","questionId":"796235373066866689","content":" π<sub>1,2,3,4,7</sub>(σ<sub>'2'>'7'Λ3=5Λ4=6</sub>(R×S))","answer":0,"chooseValue":"B"},{"id":"796235374140608513","questionId":"796235373066866689","content":" σ<sub>'2'>'7'</sub>(R×S)","answer":0,"chooseValue":"C"},{"id":"796235374169968641","questionId":"796235373066866689","content":" π<sub>1,2,3,4,7</sub>(σ<sub>2>7Λ3=5Λ4=6</sub>(R×S))","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234661025042433","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>若要将部门表Demp中name列的修改权限赋予用户Ming,并允许Ming将该限授予他人,实现的SQL语句如下:<br> GRANT(44)ON TABLE Demp TO Ming(45)。","analyze":"本题考查对标准SQL授权语句的掌握。<br> 标准SQL中授权的语句格式如下:<br> <img width=\"548\" height=\"56\" src=\"https://image.chaiding.com/ruankao/7a3a612f54be33a2723bd1c73a6278a7.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> 若在授权时指定了WITH GRANT OPTION,那么获得了权限的用户还可以将权限赋给其他用户。不同类型的操作对象有不同的操作权限,常见的操作权限如表所示。<br> 表常见的操作权限<br> <img width=\"552\" height=\"160\" src=\"https://image.chaiding.com/ruankao/bc4f8780f6d308ae3d280162f0cb709f.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> 按试题要求,是要将修改属性列name权限给用户Ming,故空(44)应填写UPDATE(name)。故空(45)应填写WITH GRANT OPTION。","multi":0,"questionType":1,"answer":"C","chooseItem":["796234662082007041"],"itemList":[{"id":"796234662002315265","questionId":"796234661025042433","content":" FOR ALL","answer":0,"chooseValue":"A"},{"id":"796234662056841217","questionId":"796234661025042433","content":" CASCADE","answer":0,"chooseValue":"B"},{"id":"796234662082007041","questionId":"796234661025042433","content":" WITH GRANT OPTION","answer":1,"chooseValue":"C"},{"id":"796234662098784257","questionId":"796234661025042433","content":" WITH CHECK OPTION","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234897701228545","title":"若某企业信息系统的应用人员分为三类:录入、处理和查询,那么用户权限管理的方案适合采用( )。","analyze":"本题考查对数据库应用系统安全策略的掌握。企业信息系统的使用人员可能很多,也可能经常变动,针对每个使用人员都创建数据库用户可能不切实际,也没有必要,而因为权限问题对关系模式修改更不可取。正确的策略是根据用户角色共享同一数据库用户,个人用户的标识和鉴别通过建立用户信息表存储,由应用程序来管理。而该类用户对数据库对象的操作权限由DBMS的授权机制管理。","multi":0,"questionType":1,"answer":"B","chooseItem":["796234898653335553"],"itemList":[{"id":"796234898632364033","questionId":"796234897701228545","content":" 针对所有人员建立用户名并授权","answer":0,"chooseValue":"A"},{"id":"796234898653335553","questionId":"796234897701228545","content":" 建立用户角色并授权","answer":1,"chooseValue":"B"},{"id":"796234898678501377","questionId":"796234897701228545","content":" 建立每类人员的视图并授权给每个人","answer":0,"chooseValue":"C"},{"id":"796234898695278593","questionId":"796234897701228545","content":" 对关系进行分解,每类人员对应一组关系","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233927084756993","title":"给定关系R(A,B,C,D,E)和关系S(D,E,F,G),对其进行自然连接运算R⋈S后其结果集的属性列为()。","analyze":"3.自然连接(Natural join)</br><br/>\n自然连接是一种特殊的等值连接</br><br/>\n两个关系中进行比较的分量必须是相同的属性组在结果中把重复的属性列去掉</br><br/>\n自然连接的含义</br><br/>\nR和S具有相同的属性组B","multi":0,"questionType":1,"answer":"B","chooseItem":["796233928351436801"],"itemList":[{"id":"796233928322076673","questionId":"796233927084756993","content":" R.A,R.B,R.C,R.D,R.E,S.D,S.E","answer":0,"chooseValue":"A"},{"id":"796233928351436801","questionId":"796233927084756993","content":" R.A,R.B,R.C,R.D,R.E,S.F,S.G","answer":1,"chooseValue":"B"},{"id":"796233928372408321","questionId":"796233927084756993","content":" R.A,R.B,R.C,R.D,R.E,S.E,S.F","answer":0,"chooseValue":"C"},{"id":"796233928389185537","questionId":"796233927084756993","content":" R.A,R.B,R.C,R.D,R.E,S.D,S.E,S.F,S.G","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235160868638721","title":"数据分析工作通常包括①~⑤五个阶段。目前,自动化程度比较低的两个阶段是( )。<br>①发现并提出问题 ②获取并清洗数据 ③按数学模型计算<br>④调整并优化模型 ⑤解释输出的结论","analyze":"本题考查应用数学(数据分析)基础知识。<br>“发现并提出问题”和“解释输出的结论”与业务领域关系更密切,更需要人的判断与经验,在人工智能尚不发达的时代,难以自动化。","multi":0,"questionType":1,"answer":"B","chooseItem":["796235161816551425"],"itemList":[{"id":"796235161787191297","questionId":"796235160868638721","content":" ①②","answer":0,"chooseValue":"A"},{"id":"796235161816551425","questionId":"796235160868638721","content":" ①⑤","answer":1,"chooseValue":"B"},{"id":"796235161841717249","questionId":"796235160868638721","content":" ③④","answer":0,"chooseValue":"C"},{"id":"796235161871077377","questionId":"796235160868638721","content":" ④⑤","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234406988632065","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>给定关系模式R&lt;U<span>,</span>F&gt;,其中:属性集U={A,B,C,D,E,G},函数依赖集F={A→BC,C→D,AE→G}。因为(42) =U,且满足最小性,所以其为R的候选码; 若将R分解为如下两个关系模式(43),则分解后的关系模式保持函数依赖。","analyze":"<img width=\"779\" height=\"416\" src=\"https://image.chaiding.com/ruankao/b53cd1ec3d5192b438d4c20f1cf858ef.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> <img width=\"776\" height=\"66\" src=\"https://image.chaiding.com/ruankao/f3eb1a47dbc6fbc69b4bbe9aa82ae53e.jpg?x-oss-process=style/ruankaodaren\" alt=\"\">","multi":0,"questionType":1,"answer":"C","chooseItem":["796234408087539713"],"itemList":[{"id":"796234408045596673","questionId":"796234406988632065","content":" <img width=\"67\" height=\"29\" src=\"https://image.chaiding.com/ruankao/d234fe78ee4a6559d2d87e0bc91a4c59.jpg?x-oss-process=style/ruankaodaren\" alt=\"\">","answer":0,"chooseValue":"A"},{"id":"796234408066568193","questionId":"796234406988632065","content":" <img width=\"65\" height=\"27\" src=\"https://image.chaiding.com/ruankao/0dbbabec5169e61f5515a5af13b8d6bc.jpg?x-oss-process=style/ruankaodaren\" alt=\"\">","answer":0,"chooseValue":"B"},{"id":"796234408087539713","questionId":"796234406988632065","content":" <img width=\"62\" height=\"29\" src=\"https://image.chaiding.com/ruankao/311eed8c0ec6cd2eef6eb9a31c1b02b5.jpg?x-oss-process=style/ruankaodaren\" alt=\"\">","answer":1,"chooseValue":"C"},{"id":"796234408108511233","questionId":"796234406988632065","content":" <img width=\"67\" height=\"25\" src=\"https://image.chaiding.com/ruankao/65c25ef185080ebc225f15d82a011c29.jpg?x-oss-process=style/ruankaodaren\" alt=\"\">","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234410109194241","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>给定关系模式R&lt;U<span>,</span>F&gt;,其中:属性集U={A,B,C,D,E,G},函数依赖集F={A→BC,C→D,AE→G}。因为(42) =U,且满足最小性,所以其为R的候选码; 若将R分解为如下两个关系模式(43),则分解后的关系模式保持函数依赖。","analyze":"<img width=\"779\" height=\"416\" src=\"https://image.chaiding.com/ruankao/b53cd1ec3d5192b438d4c20f1cf858ef.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> <img width=\"776\" height=\"66\" src=\"https://image.chaiding.com/ruankao/f3eb1a47dbc6fbc69b4bbe9aa82ae53e.jpg?x-oss-process=style/ruankaodaren\" alt=\"\">","multi":0,"questionType":1,"answer":"D","chooseItem":["796234411086467073"],"itemList":[{"id":"796234411006775297","questionId":"796234410109194241","content":" <img width=\"219\" height=\"26\" src=\"https://image.chaiding.com/ruankao/ba6704b5769c4f5782194431e69b2402.jpg?x-oss-process=style/ruankaodaren\" alt=\"\">","answer":0,"chooseValue":"A"},{"id":"796234411036135425","questionId":"796234410109194241","content":" <img width=\"232\" height=\"23\" src=\"https://image.chaiding.com/ruankao/71fe7e3f3137ad3ed542ca8072bc43ea.jpg?x-oss-process=style/ruankaodaren\" alt=\"\">","answer":0,"chooseValue":"B"},{"id":"796234411061301249","questionId":"796234410109194241","content":" <img width=\"213\" height=\"23\" src=\"https://image.chaiding.com/ruankao/1ecef625ddda4854ac286769138bf1b7.jpg?x-oss-process=style/ruankaodaren\" alt=\"\">","answer":0,"chooseValue":"C"},{"id":"796234411086467073","questionId":"796234410109194241","content":" <img width=\"229\" height=\"25\" src=\"https://image.chaiding.com/ruankao/99f58ff9418958abbad08a27fc478bb7.jpg?x-oss-process=style/ruankaodaren\" alt=\"\">","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233933753700353","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>给定关系模式R(U,F),U={A<sub>1</sub>,A<sub>2</sub>,A<span style=\"font-size:10px;\"><sub>3</sub></span>,A<sub>4</sub>}, F={A<sub>1</sub>→A<sub>2</sub>A<sub>3</sub>, A<sub>2</sub>A<sub>3</sub>→A<sub>4</sub><span></span>},那么在关系R中(43)。以下说法错误的是(44)。","analyze":"根据函数依赖的推导规则可知,对于关系模式R(U,F),如果X→Y,则XY也是候选关键字。因此,要找出R中的候选关键字,需要判断每个属性组合是否能够唯一地确定一条记录。\n<br>\n根据题意,我们有U={A1,A2,A3,A4},F={A1→A2A3, A2A3→A4},现在来分析每个属性组合:\n<br>\n{A1}:由于A1→A2A3,因此{A1}+={A1,A2,A3},且{A1,A2,A3}能唯一地确定记录。因此,{A1}是一个候选关键字。<br>\n{A2,A3}:由于A1→A2A3,因此{A2,A3}+={A1,A2,A3},且{A1,A2,A3}能唯一地确定记录。因此,{A2,A3}是一个候选关键字。<br>\n{A2,A4}:由于A1→A2A3,但A2、A4之间没有函数依赖关系,因此{A2,A4}不能唯一地确定记录,不是候选关键字。<br>\n{A3,A4}:与{A2,A4}的情况类似,{A3,A4}也不是候选关键字。<br>\n因此,R中只有一个候选关键字,即{A1}。选项 A 是正确答案。\n\n根据函数依赖“A2A3→A4”可以推导出“A2→A4”和“A3→A4”,但是反过来则不成立,不能通过已知的函数依赖推导出“A2A3→A4”。\n<br>\n因此,答案是 C。\n","multi":0,"questionType":1,"answer":"C","chooseItem":["796233934756139009"],"itemList":[{"id":"796233934722584577","questionId":"796233933753700353","content":" 根据F中“A<sub>1</sub>→A<sub>2</sub>A<sub>3</sub>”,可以得出“A<sub>1</sub>→A<sub>2</sub>”","answer":0,"chooseValue":"A"},{"id":"796233934739361793","questionId":"796233933753700353","content":" 根据F中“A<sub>1</sub>→A<sub>2</sub>A<sub>3</sub>”,可以得出“A<sub>1</sub>→A<sub>3</sub>”","answer":0,"chooseValue":"B"},{"id":"796233934756139009","questionId":"796233933753700353","content":" 根据F中“A<sub>2</sub>A<sub>3</sub>→A<sub>4</sub><span></span>”,可以得出“A<sub>2</sub>→A<sub>4</sub><span></span>,A<sub>3</sub>→A<sub>4</sub><span></span>”","answer":1,"chooseValue":"C"},{"id":"796233934777110529","questionId":"796233933753700353","content":" 根据F中“A<sub>1</sub>→A<sub>2</sub>A<sub>3</sub>,A<sub>2</sub>A<sub>3</sub>→A<sub>4</sub><span></span>”,可以得出“A<sub>1</sub>→A<sub>4</sub><span></span>”","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233930633138177","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>给定关系模式R(U,F),U={A<sub>1</sub>,A<sub>2</sub>,A<sub>3</sub>,A<sub>4</sub>}, F={A<sub>1</sub>→A<sub>2</sub>A<sub>3</sub>, A<sub>2</sub>A<sub>3</sub>→A<sub>4</sub>},那么在关系R中(43)。以下说法错误的是(44)。","analyze":"根据函数依赖的推导规则可知,对于关系模式R(U,F),如果X→Y,则XY也是候选关键字。因此,要找出R中的候选关键字,需要判断每个属性组合是否能够唯一地确定一条记录。\n<br>\n根据题意,我们有U={A1,A2,A3,A4},F={A1→A2A3, A2A3→A4},现在来分析每个属性组合:\n<br>\n{A1}:由于A1→A2A3,因此{A1}+={A1,A2,A3},且{A1,A2,A3}能唯一地确定记录。因此,{A1}是一个候选关键字。<br>\n{A2,A3}:由于A1→A2A3,因此{A2,A3}+={A1,A2,A3},且{A1,A2,A3}能唯一地确定记录。因此,{A2,A3}是一个候选关键字。<br>\n{A2,A4}:由于A1→A2A3,但A2、A4之间没有函数依赖关系,因此{A2,A4}不能唯一地确定记录,不是候选关键字。<br>\n{A3,A4}:与{A2,A4}的情况类似,{A3,A4}也不是候选关键字。<br>\n因此,R中只有一个候选关键字,即{A1}。选项 A 是正确答案。<br><br><br>\n\n根据函数依赖“A2A3→A4”可以推导出“A2→A4”和“A3→A4”,但是反过来则不成立,不能通过已知的函数依赖推导出“A2A3→A4”。\n<br>\n因此,答案是 C。\n","multi":0,"questionType":1,"answer":"A","chooseItem":["796233931711074305"],"itemList":[{"id":"796233931711074305","questionId":"796233930633138177","content":" 有1个候选关键字A<sub>1</sub>","answer":1,"chooseValue":"A"},{"id":"796233931727851521","questionId":"796233930633138177","content":" 有1个候选关键字A<sub>2</sub><span></span>A<sub>3</sub><span></span>","answer":0,"chooseValue":"B"},{"id":"796233931748823041","questionId":"796233930633138177","content":" 有2个候选关键字A<sub>2</sub><span></span>和A<sub>3</sub><span></span>","answer":0,"chooseValue":"C"},{"id":"796233931769794561","questionId":"796233930633138177","content":" 有2个候选关键字A<sub>1</sub><span></span>和A<sub>3</sub><span></span>","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234403515748353","title":"若事务T<sub>1</sub>对数据D<sub>1</sub>已加排它锁,事务T<sub>2</sub>对数据D<sub>2</sub>已加共享锁,那么(41)。","analyze":"本题考查数据库并发控制方面的基础知识。<br> 在多用户共享的系统中,许多用户可能同时对同一数据进行操作,可能带来的问题是数据的不一致性。为了解决这一问题,数据库系统必须控制事务的并发执行,保证数据处于一致的状态,在并发控制中引入两种锁:排它锁(Exclusive Locks,简称X锁)和共享锁(Share Locks,简称S锁)。<br> 排它锁又称为写锁,用于对数据进行写操作时进行锁定。如果事务T对数据A加上X锁,就只允许事务T读取和修改数据A,其他事务对数据A不能再加任何锁,因而也不能读取和修改数据A,直到事务T释放A上的锁。<br> 共享锁又称为读锁,用于对数据进行读操作时进行锁定。如果事务T对数据A加上S锁,事务T就只能读数据A但不可以修改,其他事务可以再对数据A加S锁来读取,只要数据A上有S锁,任何事务都只能再对其加S锁读取而不能加X锁修改。","multi":0,"questionType":1,"answer":"D","chooseItem":["796234404803399681"],"itemList":[{"id":"796234404711124993","questionId":"796234403515748353","content":" 事务T<sub>1</sub>对数据D<sub>2</sub>加共享锁成功,加排它锁失败;事务T<sub>2</sub>对数据D<sub>1</sub>加共享锁成功、加排它锁失败","answer":0,"chooseValue":"A"},{"id":"796234404740485121","questionId":"796234403515748353","content":" 事务T<sub>1</sub>对数据D<sub>2</sub>加排它锁和共享锁都失败:事务T<sub>2</sub>对数据D<sub>1</sub>加共享锁成功、加排它锁失败","answer":0,"chooseValue":"B"},{"id":"796234404774039553","questionId":"796234403515748353","content":" 事务T<sub>1</sub>对数据D<sub>2</sub>加共享锁失败,加排它锁成功;事务T<sub>2</sub>对数据D<sub>1</sub>加共享锁成功、加排它锁失败","answer":0,"chooseValue":"C"},{"id":"796234404803399681","questionId":"796234403515748353","content":" 事务T<sub>1</sub>对数据D<sub>2</sub>加共享锁成功,加排它锁失败;事务T<sub>2</sub>对数据D<sub>1</sub>加共享锁和排它锁都失败","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233923431518209","title":"高校信息系统采用分布式数据库系统,该系统中“当某一场地故障时,系统可以使用其它场地上的副本而不至于使整个系统瘫痪”和“数据在不同场地上的存储”分别称为分布式数据库的()。","analyze":"高校信息系统采用分布式数据库系统,其中“当某一场地故障时,系统可以使用其它场地上的副本而不至于使整个系统瘫痪”是指分布式数据库系统的容错性,而“数据在不同场地上的存储”则是指分布式数据库系统的分布性。因此,选项D. 可用性和分布性 是正确的。共享性和分布性不是分布式数据库系统的特征,自治性和可用性虽然是分布式数据库系统的属性之一,但并不能完全概括题目中所述内容。","multi":0,"questionType":1,"answer":"D","chooseItem":["796233924526231553"],"itemList":[{"id":"796233924425568257","questionId":"796233923431518209","content":" 共享性和分布性","answer":0,"chooseValue":"A"},{"id":"796233924467511297","questionId":"796233923431518209","content":" 自治性和分布性","answer":0,"chooseValue":"B"},{"id":"796233924501065729","questionId":"796233923431518209","content":" 自治性和可用性","answer":0,"chooseValue":"C"},{"id":"796233924526231553","questionId":"796233923431518209","content":" 可用性和分布性","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null}]}}