武汉理工大学数据库系统原理总复习题(完整版含答案)
数据库系统概论复习题及答案

数据库系统概论复习题及答案(共65页)--本页仅作为文档封面,使用时请直接删除即可----内页可以根据需求调整合适字体及大小--第一学期期末考试试卷和答案试卷代码:03115 授课课时:96课程名称:数据库系统原理A 适用对象:本科选课班一、选择题(从下列各题四个答案中选出一个正确答案,每小题1分,共10分)1、在数据库技术发展的几个阶段中,数据独立性最高的是__A___阶段。
A、数据库系统B、文件系统C、人工管理D、数据项管理2、在SQL的SELECT语句中,与选择运算对应的命令动词是__C___。
A、SELECTB、FROMC、WHERED、ORDER BY3、在数据库中,下列说法_A__是不正确的A、数据库避免了一切数据的重复B、若系统是完全可以控制的,则系统可确保更新是的一致性C、数据可以共享D、数据库减少了冗余4、在数据库系统中,模式/外模式映像用于解决数据的_C__A、结构独立性B、物理独立性C、逻辑独立性D、分布独立性5、关系代数的5种基本运算是__D_。
A、并、差、选择、投影、自然连接B、并、差、交、选择、投影C、并、差、交、选择、笛卡尔积D、并、差、选择、投影、笛卡尔积6、在SQL语句中,谓词“EXISTS”的含义是_B___。
A、全称量词B、存在量词C、自然连接--在连接条件中使用等于(=)运算符比较被连接列的列值,但它使用选择列表指出查询结果集合中所包括的列,并删除连接表中的重复列D、等值连接--在连接条件中使用等于号(=)运算符比较被连接列的列值,其查询结果中列出被连接表中的所有列,包括其中的重复列7、规范化过程主要为克服数据库逻辑结构中的插入异常、删除异常、更新异常以及_C__的缺陷A、数据不一致性B、结构不合理C、冗余度大D、数据丢失8、数据库数据的正确性和相容性是数据库的__B____。
A、安全性B、可维护性C、完整性D、并发控制9、数据库三级模式体系结构主要的目标是确保数据库的_B__。
武汉理工大学数据库答案汇总

第一章:1.1What is the purpose of a database?数据库的目的The purpose of a database is to help people track(跟踪监测) of things.1.2What is the most commonly used type of database?最常见的数据库类型The most commonly used type of database is the relational database.1.3Give an example of two related tables other than the example used in this book.Use the STUDENT and GRADE tables in Figure 1-3 as an example pattern(模式) for your tables. Name the tables and columns using the conventions in this book.给一个关联表的例子,参考1-3Answer:1.4For the tables you created in Review Question 1.3, what are the primary keys ofeach table? Do you think that any of these primary keys could be surrogate(代理) keys?每张表的主码?这些主码可以是代理码吗?Query1: StudentNumberQuery2: StudentNumber, RoomNumberYes, both.1.5 Explain how the two tables you provided in Review Question 1.3 are related. Which table contains the foreign key, and what is the foreign key?The primary key of Q1 were added to the Q2 with a primary key of StudentNumber to uniquely identify each row.In Q2 StudentNumber and RoomNumber each now serves as a foreign key.1.6 Show your two tables from Review Question 1.3 without the columns that represent the relationships. Explain how the value of your two tables is diminished(减少) without the relationships.在你的表中去掉代表关系的列,解释没有关系的表的value如何减少?1.7 Define the terms data and information. Explain how the two terms differ.定义术语data和information,解释它俩的不同。
(完整版)数据库系统原理(含答案)

数据库系统原理自测题(2)一、单项选择题1.数据库物理存储方式的描述称为【B】A.外模式B.内模式C.概念模式D.逻辑模式2.在下面给出的内容中,不属于DBA职责的是【A】A.定义概念模式B.修改模式结构C.编写应用程序D.编写完整行规则3.用户涉及的逻辑结构用描述【C】A.模式B.存储模式C.概念模型D.逻辑模式4.数据库在磁盘上的基本组织形式是【B】A.DB B.文件 C.二维表 D.系统目录5.在DBS中,最接近于物理存储设备一级的结构,称为【D】A.外模式B.概念模式C.用户模式D.内模式6.从模块结构考察,DBMS由两大部分组成:【B】A.查询处理器和文件管理器B.查询处理器和存储管理器C.数据库编译器和存储管理器D.数据库编译器和缓冲区管理器7.设W=R⋈S,且W、R、S的属性个数分别为w、r和s,那么三者之间应满足【A】A.w≤r+s B.w<r+s C.w≥r+s D.w>r+s8.数据库系统的体系结构是数据库系统的总体框架,一般来说数据库系统应具有三级模式体系结构,它们是【A】A.外模式、逻辑模式和内模式B.内模式、用户模式和外模式C.内模式、子模式和概念模式D.子模式、模式和概念模式9.ER图是表示概念模型的有效工具之一,在ER图中的菱形框表示【A】A.联系B.实体C.实体的属性D.联系的属性10.数据库管理系统中数据操纵语言DML所事项的操作一般包括【A】A.查询、插入、修改、删除B.排序、授权、删除C.建立、插入、修改、排序D.建立、授权、修改11.设有关系R(A,B,C)和关系S(B,C,D),那么与R⋈S等价的关系代数表达式是【C】A.π1,2,3,4(σ2=1∧3=2(R×S))B.π1,2,3,6(σ2=1∧3=2(R×S))C.π1,2,3,6(σ2=4∧3=5(R×S))D.π1,2,3,4(σ2=4∧3=5(R×S))12.在关系模式R中,函数依赖X→Y的语义是【B】A.在R的某一关系中,若两个元组的X值相等,则Y值也相等B.在R的每一关系中,若两个元组的X值相等,则Y值也相等C.在R的某一关系中,Y值应与X值相等D.在R的每一关系中,Y值应与X值相等13.设有关系模式R(A,B,C,D),R上成立的FD集F={A→C,B→C},则属性集BD 的闭包(BD)+为【B】A.BD B.BCD C.ABD D.ABCD14.有10个实体类型,并且它们之间存在着10个不同的二元联系,其中2个是1:1联系类型,3个是1:N联系类型,5个是M:N联系类型,那么根据转换规则,这个ER结构转换成的关系模式有【B】A.13个B.15个 C.18个 D.20个15.关系模式R分解成数据库模式ρ的一个优点是【D】A.数据分散存储在多个关系中B.数据容易恢复C.提高了查询速度D.存储悬挂元组16.事务并发执行时,每个事务不必关心其他事务,如同在单用户环境下执行一样,这个性质称为事务的【D】A.持久性B.一致性 C.孤立性 D.隔离性17.用户或应用程序使用数据库的方式称为【B】A.封锁B.权限 C.口令D.事务18.常用的关系运算是关系代数和。
数据库系统原理复习 题集附答案

数据库系统原理复习题集附答案数据库系统是计算机科学和技术中的一个重要领域,它涉及数据的组织、存储和管理。
在数据库系统原理的学习中,复习一些题目并附带答案是非常有效的学习方法。
下面是一些数据库系统原理的复习题及其答案,希望对大家的学习有所帮助。
1. 数据库系统的三级模式分别是什么?(5分)答案:数据库系统的三级模式包括外模式(用户级模式)、概念模式(逻辑模式)和内模式(物理模式)。
2. 数据库是什么?数据库管理系统是什么?(5分)答案:数据库是一个按照一定格式存储、组织和管理数据的集合。
数据库管理系统(DBMS)是对数据库进行管理的软件系统,它提供数据定义、数据操作和数据控制等功能。
3. 数据库系统的优点有哪些?(10分)答案:数据库系统的优点包括数据共享、数据独立性、数据安全性、数据一致性、数据可靠性和数据备份与恢复等。
4. 数据库系统的完整性约束有哪些?(10分)答案:数据库系统的完整性约束包括实体完整性约束、参照完整性约束和用户定义完整性约束。
5. 请简要解释事务的概念。
(10分)答案:事务是数据库操作的一个基本单位,它是一个不可分割的工作单位。
事务应具有四个特性,即原子性、一致性、隔离性和持久性。
6. 请解释ACID是什么意思。
(10分)答案:ACID是数据库事务的四个特性的缩写,即原子性(Atomicity)、一致性(Consistency)、隔离性(Isolation)和持久性(Durability)。
7. 请简要介绍数据库的并发控制。
(15分)答案:并发控制是数据库系统中保证事务并发执行的一种机制。
常用的并发控制方法包括锁定和时间戳。
8. 请解释关系数据库的概念。
(10分)答案:关系数据库是一种基于关系模型的数据库,以表(关系)为单位存储数据。
9. 数据库中的主键和外键有什么作用?(10分)答案:主键用于唯一标识表中的每一行数据,外键用于建立表之间的关系,实现数据的引用和关联。
10. 请解释数据库的范式。
数据库系统原理自考试题及答案

数据库系统原理自考试题及答案一、单项选择题(本大题共15小题,每题2分,共30分)在每题列出的四个备选项中只有一个是符合题目要求的,请将其代码填写在题后的括号内。
错选、多项选择或采选均无分。
1.数据库的存储设备和存取方法变化不影响整体规律布局的特点,称为()A.物理布局独立性B.物理数据独立性C.规律布局独立性D.规律数据独立性答案:B2.DBS中,“联机存储器”是指()A.内存和外存B.内存和快闪存C.磁盘和光盘D.快闪存和磁盘答案:D3.设有关系模式EMP(职工号,姓名,年龄,技能)。
假设职工号唯一,每个职工有多项技能,那么EMP表的主键是()姓名,技能技能职工号,技能答案:DA.B.C.D.职工号4.以下选项中,不正确的是()A.R= (R-S) U (RNS)B.R-S=R- (RNS)C.RNS=S-(R-S)D.RNS=S-(S-R)答案:C5.在SQL语言中,用于测试列值非空的短语是()A. IS NOT EMPTYB. IS NOT NULLC. NOT UNIQUED.NOT EXISTS6.假设关系模式R上有函数凭借AB→C和A→C,那么R 中存在()A.完全凭借B.局部凭借C.传递凭借D.多值凭借答案:B7.设有关系模式R(A,B,C,D)上的函数凭借集F={AB →CD,C→A},那么R最高属于()A. 2NFB. 3NFC. BCNFD. 4NF答案:B8.ABCD答案:BA.B.C.D.9.在需求分析阶段,数据字典是对系统中()A.数据的描述B.处理的描述C.功能的描述D.数据与处理关系的描述答案:A10.扩展的ER模型中,子类与超类之间存在着()A.相容性B.一致性C.继承性D.完整性答案:C11.某公司经销多种产品,每名业务员可推销多种产品,且每种产品由多名业务员推销,那么业务员与产品之间的联系是()A.一对一B.一对多C.多对多D.多对一答案:C12.设关系模式R(A,B,C,D)上的函数凭借集F={AD→B,C→D},F在属性集ABC上的投影IIABC(F)为()A. {AD→B, C→D}B.{C→B}C.{AB→C}D.{AC→B}答案:D13.在对象关系模型中,若属性为不同类型元素的有序集合,那么称该类型为()A.布局类型B.数组类型C.集合类型D.组合类型答案:A14.把对关系SC的属性GRADE的修改权授予用户ZHAO的SQL语句是()A.GRANT GRADE ON SC TO ZHAOB.B. GRANT UPDATE ON SC TO ZHAOC.GRANT UPDATE (GRADE) ON SC TO ZHAOD.GRANT UPDATE ON SC (GRADE) TO ZHAO答案:C。
数据库系统原理练习题库(附参考答案)

数据库系统原理练习题库(附参考答案)一、单选题(共100题,每题1分,共100分)1.属于数据库结构设计阶段的是A、程序设计B、功能设计C、事务设计D、逻辑结构设计正确答案:D2.人工管理阶段,计算机主要应用于A、数据集成B、科学计算C、过程控制D、故障恢复正确答案:B3.下列关于数据控制语言的说法中,正确的是A、REVOKE语句用于授予权限B、GRANT语句用于收回权限C、数据控制语言主要用于数据执行流程管理D、数据控制语言包括的主要SQL语句是GRANT和REVOKE正确答案:D4.下列属于关联分析算法的是A、AprioriB、GMMC、RedisD、HBase正确答案:A5.同一数据被反复存储的情况是A、删除异常B、插入异常C、更新异常D、数据冗余正确答案:D6.产生数据不一致的主要原因是并发操作破坏了事务的A、持续性B、一致性C、原子性D、隔离性正确答案:D7.DBMS提供【】来严格地定义模式。
A、模式描述语言B、子模式描述语言C、内模式描述语言D、程序设计语言正确答案:A8.关系数据库以【】作为数据的逻辑模型。
A、二维表B、关系C、关系模型D、数据库正确答案:C9.关于调用存储过程的说法,错误的是A、可以从交互式界面调用B、可以使用CALL语句来调用存储过程C、可以由嵌入式SQL调用D、不是所有的SQL接口都能调用存储过程正确答案:D10.可唯一标识实体的属性集称为A、键B、实体型C、域D、属性正确答案:A11.目的是为可实际运行的应用程序设计提供依据与指导,并作为设计评价的基础的是A、设计评价B、编制应用程序设计说明C、模型转换D、子模式设计正确答案:B12.在使用游标的过程中,需要注意的事项不包括A、游标不能单独在查询操作中使用B、在一个BEGIN···END语句块中每一个游标的名字并不是唯一的C、游标是被SELECT语句检索出来的结果集D、在存储过程或存储函数中可以定义多个游标正确答案:B13.对关系的描述不正确的是A、关系中的元组次序可交换B、关系可以嵌套定义C、关系是一张二维表D、关系是一个集合正确答案:B14.下列关于MySQL的说法中,正确的是A、在MySQL中,一个关系对应多个基本表B、在MySQL中,一个或多个基本表对应一个存储文件C、在MySQL中,一个表只能有一个索引D、在MySQL中,索引不能存放在存储文件中正确答案:B15.关系数据库是以【】的形式组织数据。
武汉理工大学数据库期末考试试题06级

武汉理⼯⼤学数据库期末考试试题06级,B) ∪,D) ∩,武汉理⼯⼤学教务处试题标准答案及评分标准⽤纸课程名称数据库与信息系统( B 卷)⼀、填空题(每空1分,共15分)1.外模式,模式(或逻辑模式)2.属性3. 1对多(1:N),多对多(M:N)4. 实体完整性约束,参照完整性约束,⾃定义完整性约束5. .mdf,.ldf6. CREATE TRIGGER7. 传递函数依赖8. R∩S9. 数据操纵语句10. DROP VIEW⼆、单项选择题(本⼤题共20⼩题,每⼩题1分,共20分)D,C,B,C,A,B,C,B,A,D,A,D,C,D,A,B,C,A,D,B三、简答题(每题5分,共15分)1.数据的4个特征:1) 数据有“型”和“值”之分。
2分2) 数据受数据类型和取值范围的约束。
1分3) 数据有定性表⽰和定量表⽰之分。
1分4) 数据应具有载体和多种表现形式。
1分2.需求分析阶段,概念结构设计阶段,逻辑结构设计阶段,数据库物理设计阶段,数据库实施阶段,数据库运⾏和维护阶段 5分3. 主语⾔通过主变量向SQL语句提供参数。
2分SQL语句当前的⼯作状态和运⾏环境数据要反馈给应⽤程序。
SQL将其执⾏信息送到通信区SQLCA 中,应⽤程序从SQLCA 中取出这些状态信息,并据此信息来控制应该执⾏的语句。
3分四、综合题1. 1)∏(课程名=‘数据库’(学⽣∞选课∞课程))5分姓名,学号2)∏姓名,学号(课程名=‘数据库’(学⽣∞选课∞课程))∩∏姓名,学号(课程名=‘数据结构’(学⽣∞选课∞课程))5分3)∏姓名,学号(学⽣)-∏姓名,学号(课程名=‘数据库’(学⽣∞选课∞课程))5分2.1)SELECT⼯程代号3分FROM供应零件WHERE供应商代码=’S1’;2)SELECT零件.零件名,供应零件.数量4分FROM零件, 供应零件WHERE零件. 零件代码=供应零件. 零件代码AND供应零件. ⼯程代码=’J2’;3)SELECT零件代号4分FROM供应商, 供应零件WHERE供应商. 供应商代码=供应零件. 供应商代码AND供应商. 所在城市=’上海’4)SELECT⼯程号码4分FROM⼯程WHERE NOT EXISTS (SELECT *FORM供应零件WHERE⼯程. ⼯程代码=供应零件. ⼯程代码AND供应商代码IN(SELECT供应商代码FROM供应商WHERE所在城市=’天津’));3. 1)CREA TE DA TABASE 学⽣选课库5分ON PRIMARY(NAME=学⽣选课库,FILENAME=’C:\msSQL\data\学⽣选课. mdf’,SIZE=4MB,MAXSIZE=6MBFILEROWHT=2MB)GO2)CREA TE TABLE 课程(课程号CHAR(10)PRIMARY KEY CLUSTERED,3分课程名V ARCHAR(50),先⾏课V ARCHAR(50))GO3)CREA TE INDEX student ind ON学⽣(学号)3分GOCREA TE INDEX class ind ON 课程(课程号)GO4)CREA TE RULE value rule AS @V ALUE IN(’男’,’⼥’)3分GOCREA TE DEFAULT 性别缺省AS’男’GO5)CREA TE TRIGGER sc insert ON选课3分FOR INSERTAS IF(SELECT COUNT(*)FROM学⽣,inserted, 课程WHERE学⽣.学号=inserted. 学号AND课程. 课程号=inserted. 课程号)=0 ROLLBACK TRANSACTIONGO6)CREA TE TRIGGER delete all ON学⽣4分FOR DELETEAS DELETEFROM选课WHERE选课. 学号=deleted. 学号GO。
(完整版)数据库原理及应用习题(有答案)

假设教学管理规定:
①一个学生可选修多门课,一门课有若干学生选修;
②一个教师可讲授多门课,一门课只有一个教师讲授;
③一个学生选修一门课,仅有一个成绩。
学生的属性有学号、学生姓名;教师的属性有教师编号,教师姓名;课程的属性有课程号、课程名。
要求:根据上述语义画出 ER 图,要求在图中画出实体的属性并注明联系的类型;
(1)检索“程军”老师所授课程的课程号(C#)和课程名(CNAME)。 (2)检索年龄大于 21 的男学生学号(S#)和姓名(SNAME)。 (3)检索至少选修“程军”老师所授全部课程的学生姓名(SNAME)。 (4)检索”李强”同学不学课程的课程号(C#)。 (5)检索至少选修两门课程的学生学号(S#)。 (6)检索全部学生都选修的课程的课程号(C#)和课程名(CNAME)。 (7)检索选修课程包含“程军”老师所授课程之一的学生学号(S#)。
4、SQL 语言具有两种使用方式,分别称为交互式 SQL 和 嵌入式 SQL 。
5、假定学生关系是 S(S#,SNAME,SEX,AGE),课程关系是 C(C#,CNAME,TEACHER),学生选课关系是
SC(S#,C#,GRADE)。要查找选修“COMPUTER”课程的“女”学生姓名,将涉及到关系 S,C,SC 。
三:简答题:
1. 什么是数据库?
答:数据库是长期存储在计算机内、有组织的、可共享的数据集合。数据库是按某种数据模型进行
组织的、存放在外存储器上,且可被多个用户同时使用。因此,数据库具有较小的冗余度,较高的数据
独立性和易扩展性。
2. 什么是数据库的数据独立性?
答:数据独立性表示应用程序与数据库中存储的数据不存在依赖关系,包括逻辑数据独立性和物理
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理工大学数据库系统原理总复习题(完整版含答案)1. Questions1.1What is the purpose of a database?ANSWER:The purpose of a database is to help people track of things.1.2What is the most commonly used type of database?ANSWER: the most commonly used type of database is the relational database.1.7 Define the terms data and information. Explain how the two terms differ.ANSWER: Data are recorded facts and numbers.we can now define information as:●Knowledge derived from data.●Data presented in a meaningful context.●Data processed by summing, ordering, averaging, grouping,comparing or other similar operations.1.10 What problem can occur when a database is processed by more than one user?ANSWER: When more than one user employs a database application, these is always the chance that one user's work may interfere with other's.1.12 What is the purpose of the largest databases at e-commerce companies such as Amazon.?ANSWER: The largest databases are those that track customer browser behavior.(用来记录用户的浏览行为的。
)1.13 How do the e-commerce companies use these databases? .ANSWER: E-commerce companies use Web activity databases to determine which items on a Web page are popular and successful and which are not.1.14 How do digital dashboard and data mining applications differ from transaction processing applications?ANSWER: Digital dashboards and other reporting systems assess past and current performance. Data mining applications predict future performance.1.15 Explain why a small database is not necessarily simpler than a large one.What are the functions of application programs?ANSWER: Supposed we have 2 company which are different in sales but have similar database. Though the difference in sale, both have the same kinds of data, about the same number of tables of data, and the same levelof complexity in data relationships. Only the amount of data varies from one to the other. Thus, although a database for a small business may be small, it is not necessarily simple.1.18 What is Structured Query Language (SQL), and why is it important?ANSWER: Structured Query Language (SQL)is an internationally recognized standard language. Because it can be understood by all commercial DBMS products, in database processing and the fact that database applications typically send SQL statements to the DBMS for processing.1.19 What does DBMS stand for?ANSWER: The database management system.1.20 What are the functions of the DBMS?ANSWER: It can be used to create, process, and administer the database.1.21 Name three vendors of DBMS products.ANSWER: IBM, Microsoft, Oracle.1.22 Define the term database.ANSWER: A database is a self-describing collection of integratedtables.1.23 Why is a database considered to be self-describing?ANSWER: A database is self-describing because it contains a description of itself. Thus, databases contain not only tables of user data, but also tables of data that describe that user data.1.24 What is metadata? How does this term pertain to a database?ANSWER: databases contain not only tables of user data, but also tables of data that describe that user data. Such descriptive data is called metadata because it is data about data.1.25 What advantage is there in storing metadata in tables?ANSWER: Because metadata is stored in tables, you can use SQL to query it. Thus, by learning how to write SQL to query user tables, you will also learn how to write SQL to query metadata.1.26 List the components of a database other than user tables and metadata.• Tables of user data• Metadata• Indexes• Store d procedures• Triggers• Security data• Backup/recovery data1.27 Is Microsoft Access a DBMS? Why or why not?ANSWER: No, Microsoft Access is not just a DBMS. Rather, it is a personal database system: a DBMS plus an application generator.Because although Microsoft Access contains a DBMS engine that creates, processes, and administers the database, it also contains form, report, and query components that are the Microsoft Access application generator1.37 List several consequences of a poorly designed database.ANSWER:-They may require application developers to write overly complex and contrived SQL to get –wanted data.-they may be difficult to adapt to new and changing requirements.-they may fail in some other way.1.38 Explain two ways that a database can be designed from existingdata.ANSWER:The first type of database design involves databases that are constructed from existing Data.A second way that databases are designed is for the development of new information systems.1.39 What is a data warehouse? What is a data mart?ANSWER: The data warehouse and data mart databases store data specifically organized for research and reporting purposes, and these data often are exported to other analytical tools, such as SAS’s Enterprise Miner, IBM's SPSS Data Modeler, or TIBCO's Spot fire Metrics.1.40 Describe the general process of designing a database for a new information system.ANSWER: First, the team creates a data model from the requirements statements and then transforms that data model into a database design.1.41 Explain two ways that databases can be redesigned.ANSWER:In the first, a database is adapted to new or changing requirements. This process sometimes is called database migration. In the migration process, tables may be created, modified, or removed; relationships may be altered; data constraints may be changed; and so forth.The second type of database redesign involves the integration of two or more databases. This type of redesign is common when adapting or removing legacy systems. It is also common for enterprise application integration, when two or more previously separate information systems are adapted to work with each other.1.42 What does the term database migration mean?ANSWER: The process of a database is adapted to new or changingrequirements.1.43 Summarize the various ways that you might work with database technology.ANSWER: In our career, we may work with database technology as either a user or as a database administrator.As a user, you may be a knowledge worker who prepares reports, mines data, and does other types of data analysis or you may be a programmer who writes applications that process the database.Alternatively, you might be a database administrator who designs, constructs, and manages the database itself. Users are primarily concerned with constructing SQL statements to get and put the data they want. Database administrators are primarily concerned with the managementof the database.1.44 What job functions does a knowledge worker perform?ANSWER: preparing reports, mining data, and doing other types of data analysis.1.45 What job functions does a database administrator perform?ANSWER: designing, constructing, and managing the database itself.1.47 What need drove the development of the first database technology?ANSWER: The need for data integration drove the development of the first database technology.1.48 What are Data Language/I and CODASYL DBTG?ANSWER:Data Language/I (DL/I) used hierarchies or trees (see Appendix G) to represent relationships.This subcommittee developed a standard data model that came to bear its name—the CODASYL DBTG model. It was an unnecessarily complicated model. This data relationship used data structures called networks.1.49 Who was E. F. Codd?ANSWER: E.F.Codd was a little-known IBM engineer published a paper in the Communications of the ACM3 in which he applied the concepts of a branch of mathematics called relational algebra to the problem of “shared data banks,” as databases were then known. The results of this work are now the relational model for databases, and all relational database DBMS products are built on this model.1.50 What were the early objections to the relational model?1.51 Name two early relational DBMS products.ANSWER: Oracle Database, DB2.1.52 What are some of the reasons for the success of Oracle Database?ANSWER:1, it would run on just about any computer and just about any operating system.2, Oracle Database had, and continues to have, an elegant and efficient internal design.1.53 Name three early personal computer DBMS products.ANSWER: dBase, R:base, Paradox.1.55 What was the purpose of OODBMS products? State two reasons that OODBMS products were not successful.ANSWER: They were designed to make it easy to store the data encapsulated in OOP objects.There were two reasons for their lack of acceptance. First, using an OODBMS required that the relational data be converted from relational format to object-oriented format. By the time OODBMS emerged, billions upon billions of bytes of data were stored in relational format in organizational databases. No company was willing to undergo the expensive travail of converting those databases to be able to use the new OODBMS. Second, object-oriented databases had no substantial advantage over relational databases for most commercial database processing. As you will see in the next chapter, SQL is not object oriented. But it works, and thousands of developers have created programs that use it. Without a demonstrable advantage over relational databases, no organization was willing to take on the task of converting their data to OODBMS format.1.56 What characteristic of HTTP was a problem for database processing applications?ANSWER: HTTP is a stateless protocol; a server receives a request from a user, processes the request, and then forgets about the user and the request. Many database interactions are multistage. A customer views products, adds one or more to a shopping cart, views more products, adds more to the shopping cart, and eventually checks out. A stateless protocol cannot be used for such applications.1.57 What is an open source DBMS product? Which of the five DBMS products that you named in answering Review Question 1.36 is historically an open source DBMS product?ANSWER: the MySQL DBMS.这套试卷没有1.36题啊。