数据库原理与应用试题库1

数据库原理与应用试题库1
数据库原理与应用试题库1

三、简答题

1.设有关系模式:学生修课管理(学号,,所在系,性别,课程号,课程名,学分,成绩)。设一名学生可以选修多门课程号,一门课程号可以被多名学生选修;一名学生有唯一的所在系,每门课程号有唯一的课程名和学分。

回答以下问题:(1)根据上述规定写出关系模式R的基本函数依赖;

(2)找出关系模式R的候选码;

(3)试问关系模式R最高已经达到第几式?为什么?

(4)将R分解成3NF模式集。

答:(1)学号(,所在系,性别)F

课程号(课程名,学分)F

(学号,课程号)成绩F

(学号,课程号)(,所在系,性别,课程号,学分)P

(2)候选码:学号,课程号

(3)存在部分函数依赖,R达到第一式

(4)Student(学号,,所在系,性别)

SC(学号,课程号,成绩)

Course(课程号,课程名,学分)

2.设有关系模式:学生表(学号,,所在系,班号,班主任,系主任)。其语义为:一名学生只在一个系的一个班学习,一个系只有一名系主任,一个班只有一名班主任,一个系可以有多个班。

回答以下问题:(1)根据上述规定写出关系模式R的基本函数依赖;

(2)找出关系模式R的候选码;

(3)试问关系模式R最高已经达到第几式?为什么?

(4)将R分解成3NF模式集。

答:(1)学号(,所在系,班号,班主任,系主任)F

班号(班主任,系主任)F

班主任系主任F

所在系系主任F

(2)候选码:学号

(3)存在传递依赖,不存在部分函数依赖,R达到第二式

(4)Student(学号,,所在系,班号)

Class(班号,班主任)

Dept(所在系,系主任)

3.设有关系模式:授课表(课程号,课程名,学分,授课教师号,教师名,授课时数)。其语义为:一门课程号有确定的课程名和学分,每名教师有确定的教师名,每门课程号可以由多名教师讲授,每名教师也可以讲授多门课程,每名教师对每门课程号有确定的授课时数。

回答以下问题:(1)根据上述规定写出关系模式R的基本函数依赖;

(2)找出关系模式R的候选码;

(3)试问关系模式R最高已经达到第几式?为什么?

(4)将R分解成3NF模式集。

答:(1)课程号(课程名,学分)F

授课教师号教师名F

(授课教师号,课程号)授课时数F

(授课教师号,课程号)(课程名,学分,教师名)P

(2)候选码:授课教师号,课程号

(3)存在部分函数依赖,R达到第一式

(4)Course(课程号,课程名,学分)

Teacher(授课教师号,教师名)

CT(课程号,授课教师号,授课时数)

4.(20分)设某图书集团有一关系模式R如下:R(书店编号,书籍编号,库存数量,部门编号,负责人)如果规定:(1)每个书店的每种书籍只在该书店的一个部门销售;

(2)每个书店的每个部门只有一个负责人;

(3)每个书店的每种书籍只有一个库存数量。

回答以下问题:(1)根据上述规定写出关系模式R的基本函数依赖;

(2)找出关系模式R的候选码;

(3)试问关系模式R最高已经达到第几式?为什么?

(4)将R分解成3NF模式集。

答:(1)有三个函数依赖:

(书店编号,书籍编号)部门编号(2分)

(书店编号,部门编号)负责人(2分)

(书店编号,书籍编号)库存数量(2分)

部门编号负责人

书籍编号库存数量

(2)R的候选码:(书店编号,书籍编号)(3分)

(3)R属于2NF。(2分)

因为R中存在着非主属性“负责人”对候选码(书店编号,书籍编号)的传递函数依赖,所以R属于2NF。(3分)(4)分解成:

R1(书店编号,书籍编号,库存数量,部门编号)(3分)

R2(书店编号,部门编号,负责人)(3分)

四、综合题

1.设有一个SPJ数据库,包括S、P、J、SPJ四个关系模式:

供应商表S(供应商代码SNO,供应商SNAME,供应商状态STATUS,供应商所在城市CITY);

零件表P(零件代码PNO,零件名PNAME,颜色COLOR,重量WEIGHT);

工程项目表J(项目代码JNO,项目名JNAME,项目所在城市CITY);

供应情况表SPJ(供应商代码SNO,零件代码PNO,项目代码JNO,供应数量QTY);

用SQL语言完成如下查询:(1)找出所有供应商的和所在城市;

(2)求供应工程J1零件P1的供应商SNO;

(3)求供应工程J1零件为红色的供应商SNO;

(4)找出所有零件的名称、颜色、重量;

(5)找出厂商供应的所有零件;

(6)找出工程项目J2使用的各种零件的名称及其数量;

(7)找出所有供应商的和所在城市;

(8)找出所有零件的名称、颜色、重量;

(9)找出使用供应商S1所供应零件的工程;

(10)找出工程项目J2使用的各种零件的名称及其数量。

(1)select SNAME,CITY from S (2分)

(2)select SNO from SPJ where JNO=’J1’ and PNO=’P1’(3分)

(3)select SNO from SPJ where SPJ.PNO=P.PNO and JNO=’J1’ and COLOR=’red’(3分)

(4)select PNAME,COLOR,WEIGHT from P; (3分)

(5)select S.SNO from S,P,SPJ where S.SNO=SPJ.SNO and SPJ.PNO=P.PNO and CITY=’’;(3分)

(6)select PNAME,WEIGHT from P,SPJ where P.PNO=SPJ.PNO and JNO=’J2’; (3分)

(7)Select sname, city1 from S ; (2分)

(8)Select Pname, color, weight from P;(2分)

(9)Select Jno from SPJ where sno=’s1’; (3分)

(10)Select P.pname , SPJ.qty from SPJ, P where SPJ.pno = P. pno and SPJ.Jno=’J2’; (3分)

2.设某商业集团数据库中有三个实体集:

商店:商店编号、商店名、地址

商品:商品编号、商品名、规格、单价

职工:职工编号、、性别、业绩

每个商店可销售多种商品,每种商品也可放在多个商店销售,每个商店销售一种商品时有月销售量;每个商店有许多职工,每个职工只能在一个商店工作,商店聘用职工有聘期和月薪。

(1)试画出E-R图,要求在图上注明属性及联系的类型;

(2)将E-R图转换成关系模型,并注明主码;

(3)根据实际情况,使用SQL创建表,包括各种约束;

(4)用SQL语句查找大于平均业绩的职工;

(5)用SQL语句创建一个业绩大于100的所有男职工信息的视图。

(1)(5分)

(2)这个E-R图可转换为4个关系模式:(8分)

商店(商店编号,商店名,地址)(2分)

职工(职工编号,,性别,业绩,商店编号,聘期,月薪) (2分)

商品(商品编号,商品名,规格,单价)(2分)

销售(商店编号,商品编号,月销售量)(2分)

(3)create shop(Sid char(3) primary key, Sname char(10), Sadd char(50)); (2分)create employee(Eid char(3) primary key, Ename char(5), Esex char(1), Each real, Sid char(3), Ere date, Esa int, foreign key (Sid) references (shop)); (2分)

create commodity(Cid char(3) primary key, Cname char(10), Csp char(10), Cpr real); (2分)create vendition(Sid char(3), Cid char(3), Vse int, primary key (Sid,Cid),

foreign key (Sid) references (shop), foreign key (Cid) references (commodity)); (2分)

(4)select Ename from employee x where Each>=(select avg(Each) from employee y where y.Each=x.Each); (2分)

(5)create view Eman(Eid, Ename, Esex, Each, Sid, Ere, Esa)

As select Eid, Ename, Esex, Each, Sid, Ere, Es from employee where Each>100 and Esex=’男’;(2分)

3.(10分)设有学生表S(SNO,SN),其中SNO为学号,SN为;

学生选课表SC(SNO,CNO,CN,G),其中CNO为课程号,CN为课程名,G为成绩,用SQL语言完成以下各题:(1)建立一个视图V-SSC(SNO,SN,CNO,CN,G),并按CNO升序排序;(5分)

(2)从视图V-SSC上查询平均成绩在90分以上的SN,CN和G。(5分)

(1)CREATE VIEW V-SSC(SNO,SN,CNO,CN,G)

AS SELECT S.SNO, S.SN, https://www.360docs.net/doc/5910928466.html,O, https://www.360docs.net/doc/5910928466.html,, SC.G

FROM S, SC

WHERE S.SNO=SC.SNO;

ORDER BY CNO; (5分)

(2)SELEC SN, CN, G

FROM V-SSC

GROUP BY SNO

HAVING AVG(G)>90; (5分)

4. (共10分)设学校数据库中有两个实体集:

学生表:学号、、班级

课程表:课程号、课程名称、教师

某学校有若干学生,每个学生可以选修多门课程,学校有若干课程供学生选修,每门课程可以供多个学生选修,要建立该学校学生选修课程的数据库,请设计:

(1)试画出E-R 图,要求在图上注明属性及联系的类型;

(2)将E-R 图转换成关系模型,并注明主码;

(2)这个E-R 图可转换为4个关系模式:(12分)

商店(商店编号,商店名,地址) (3分)

职工(职工编号,,性别,业绩,商店编号,聘期,月薪) (3分)

商品(商品编号,商品名,规格,单价) (3分)

5.(10分)有“学生选课系统”数据库,学生选课的关系模式为:

学生(学号,,性别,年龄,所在系)

课程(课程号,课程名,先行课)

选课(学号,课程号,成绩)

根据所给系统,用关系代数运算完成下面查询。

(1)查询年龄小于20岁的学生。

(2)查询学生的和所在系,即求“学生”关系中学生和所在系两个属性上的投影。

(3)查询选修了2号课程的学生学号。

(4)查询选修了全部课程的学生的学号和。

6、(10分)设有学生表S (SNO ,SN ,SA ),其中SNO 为学号,SN 为,SA 为年龄;

学生选课表SC (SNO ,CNO ,CN ,G ),其中CNO 为课程号,CN 为课程名,G 为成绩,用SQL 语言完成以下各题:(1)查询所有年龄在20岁以下的学生及年龄。(5分)

(2)查询选修了2号课程且成绩在90分以上的所有学生的学号及。(5分)

(1)select sn,sa _______(2分)

from s ———(1分)

where sa<20;(2分)

(2)select s.sno,smame from s,sc_______(2分)

Where s.sno=sc.sno and https://www.360docs.net/doc/5910928466.html,o=’2’ and sc.g>90;————(3分)

7.设有一个SPJ 数据库,包括S 、P 、J 、SPJ 四个关系模式:

S (SNO,SNAME,STATUS,CITY )

P (PNO,PNAME,COLOR,WEIGHT )

J (JNO,JNAME,CITY )

SPJ (SNO,PNO,JNO,QTY )

供应商表S 由供应商代码(SNO )、供应商(SNAME )、供应商状态(STATUS )、供应商所在城市(CITY )组成; (2)这个E-R 图可转换为3个关系模式:(6分) 课程表(学号,姓名,班级)

选修(学号,课程号)

课程表(课程号,课程名称,教师)

零件表P由零件代码(PNO)、零件名(PNAME)、颜色(COLOR)、重量(WEIGHT)组成;

工程向目标J由工程项目代码(JNO)、工程项目名(JNAME)、工程项目所在城市(CITY)组成;

供应情况表SPJ由供应商代码(SNO)、零件代码(PNO)、工程项目代码(JNO)、供应数量(QTY)组成,表示某种供应商供应某种零件给某工程项目的数量为QTY。

今有若干数据如下:

(省略四图)

试用SQL完成如下查询:

建立题目所述的四个表,并输入数据;

求供应工程J1零件的供应商SNO;

求供应工程J1零件P1的供应商SNO;

求供应工程J1零件为红色的供应商SNO;

求没有使用天津供应商生产的红色零件的工程号JNO;(暂时不做)

求至少用了供应商S1所供应的全部零件的工程号JNO;(暂时不做)

找出所有供应商的和所在城市;

找出所有零件的名称、颜色、重量;

找出所有使用供应商S1所供应零件的工程;

找出工程项目J2使用的各种零件的名称及其数量;

找出厂商供应的所有零件;

找出使用产的零件的工程名称;

找出没有使用天津产的零件的工程;

把全部红色零件的颜色改为蓝色;

由S5供给J4的零件P6改为由S3供应,请做出必要的修改;

从供应商关系中删除S2的记录,并从供应情况关系中删除相应的记录;

请将(S2,J6,P4,200)插入供应情况关系;

答案:

Create table ;

Select sno from s,spj where s.sno=spj.sno and jno=’j1’;

Select sno from s,spj where s.sno=spj.sno and jno=’j1’ and pno=’p1’;

Select sno from s,spj,p where s.sno=spj.sno and spj.pno=p.pno and jno=’j1’ and color=’红’;

Select sname,city from s;

Select pname,color,weight from p;

Select jno from j,spj where j.jno=spj.jno and sno=’s1’;

Select pname,qty from p,spj where p.pno=spj.pno and jno=’j2’;

Select pno from p,spj,s where p.pno=spj.pno and spj.sno=s.sno and city=’’;

Select jname from jno where city=’’;

Select jno from j,spj,s where j.jno=spj.jno and spj.sno=s.sno and s.city <>’天津’;

Update p set color=’蓝’ where color=’红’;

8. 设有一个学生选课数据库,包括Student,SC,Course 三个关系模式:

Student(Sno,Sname,Ssex,Sage,Sdept)

SC(Sno,Cno,Grade)

Course(Cno,Cname,Ccredit,Semester)

试用SQL完成如下查询:

查询SC表中的全部数据;

查询计算机系学生的和年龄;

查询成绩在70-80分的学生的学号、课程号和成绩;

查询计算机系年龄在18-20岁的男学生的和年龄;

查询C001课程号的最高分;

查询计算机系学生的最大年龄和最小年龄;

统计每个系的学生人数;

统计每门课程号的选课人数和考试最高分;

统计每个学生的选课门数和考试总成绩,并按选课门数升序显示结果;

查询总成绩超过200分的学生,要求列出其学号和总成绩;

查询选修C002课程的学生和所在系;

查询成绩80分以上的学生、课程号和成绩,并按成绩降序排列结果;

查询哪些课程号没有学生选修,要求列出课程号和课程名;

查询计算机系哪些学生没有选课,列出学生;

查询选修C001课程号的学生和所在系;

查询通信工程系成绩在80分以上的学生学号和;

查询计算机系考试成绩最高的学生。

查询年龄最大的男学生的和年龄;

查询C001课程号的考试成绩高于该课程号平均成绩的学生学号和成绩;

创建容为学生学号、、所在系、课程号、课程名、课程学分的视图;

创建容为学生的学号、、选修课程名和考试成绩的视图;创建容为统计每个学生的选课门数的视图,要求列出学生学号和选课门数;

创建容为每个学生的选课总学分的视图,要求列出学生学号和总学分(说明:考试成绩超过60才能获得此课程的学分)

删除选课成绩小于50分的学生的选课记录;

将所有选修C001课程的学生的成绩加10分;

将计算机所有选修“数据库原理及应用”课程的学生成绩

加10分。

答案:

Select * from sc;

Select sname,sage from student where sdept=’计算机系’;

Select sno,course,grade from sc where grade between 70 and 80;

Select sname,sage from student where sdept=’计算机系’ and ssex=’男’ and sage between 18 and 20;Select max(grade) 最高分 from sc whereo=’C001’;Select max(sage) 最大年龄,min(sage) 最小年龄 from student where sdetp=’计算机系’;

Select sdept,count(*) 学生人数 from student group by sdept;

Selecto,count(*) 选课人数, max(grade) 最高分 from sc group byo;

Select sno,count(*) 选课门数, sum(grade) 总成绩from sc group by sno order by count(*);

Select sno,sum(grade) 总成绩 from sc group by sno having sum(grade)>200;

Select sname,sdept from student,sc where student.sno=sc.sno ando=’C002’;

Select sname,cno,grade from student,sc where student.sno=sc.sno and grade>80 order by grade desc;Select

Select

Select

Select sno,sname from student,sc where student.sno=sc.sno and sdept=’通信工程系’ and grade>80;

Select

Select

Select sno,grade from sc where grade>(select avg(grade) from sc whereo=’c001’) ando=’c001’;

Create view view1 as select sno,sname,sdept,cno,cname,ccredit from student,sc,course where student.sno=sc.sno and https://www.360docs.net/doc/5910928466.html,o=https://www.360docs.net/doc/5910928466.html,o;

Create view view2 as select sno,sname,cname,grade from student,sc,course where student.sno=sc.sno and https://www.360docs.net/doc/5910928466.html,o=https://www.360docs.net/doc/5910928466.html,o;

大学英语四级模拟试题四(附含答案解析)

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