数据库第三章作业

数据库第三章作业

基本表：

$S(SNO,SNAME,AGE,SEX)$

$SC(SNO,CNO,GRADE)$

$C(CNO,CNAME,TEACHER)$

问题一：使用SQL的查询语句表达下列查询：

1. 检索Liu老师所教授课程的课程号和课程名

   select CNO, CNAME
   from C
   where TEACHER = 'Liu'
   

2. 检索年龄大于23岁的男学生的学号和姓名

   select SNO, SNAME
   from S
   where AGE > 23 and SEX = '男';
   

3. 检索学号为S3学生所学课程的课程名与任课教师名

   select CNAME, TEACHER
   from C, SC
   where C.CNO = SC.CNO and SC.SNO = 'S3';
   

4. 检索至少选修Liu老师所授课程中一门课程的女同学的姓名

   select distinct NAME
   from S, SC, C
   where S.SNO = SC.SNO and SC.CON = C.CON and C.TEACHER = 'Liu' and S.SEX = '女';
   

5. 检索Wang同学不学的课程的课程号

   (select distinct CNO
    from S, SC
    where S.SNO = SC.SNO and S.SNAME = 'Wang')
   except 
   (select distinct CNO
    from S, SC, C
    where S.SNO = SC.SNO and SC.CON = C.CON and C.TEACHER = 'Wang';)
   

6. 检索至少选修两门课程的学生学号

   select distinct SNO
   from SC as X
   where SNO in (select SNO
        		  from SC as Y
        		  where X.SNO = Y.SNO and X.CNO != Y.CNO);
   

7. 检索全部学生都选修的课程的课程号和课程名

   select CNO,CNAME
   from C
   where not exists ( (select SNO 
        				from S)
       				except
       			   (select distinct SNO
       				from SC
       				where SC.CNO = C.CNO) );
   

8. 检索选修课程包含Liu老师所授课程的学生学号

   select SNO
   from S
   where exists (select SNO 
                 from SC, C
                 where SC.CNO = C.CNO and SC.SNO = S.SNO and C.TEACHER = 'Liu');
   

问题二：试用SQL查询语句表达下列对三个基本表S、SC、C的查询：

1. 在表C中统计开设课程的教师人数

   select count(distinct TEACHER)
   from SC;
   

2. 求选修C4课程的女学生的平均年龄

   select avg(AGE)
   from S, SC, C
   where S.SNO = SC.SNO and SC.CNO = C.CNO and C.CNAME = 'C4' and S.SEX = '女';
   

3. 求Shu老师所授课程的每门课程的平均成绩

   select CNO, avg(GRADE)
   from SC, C
   where SC.CNO = C.CNO and C.TEACHER = 'Shu'
   group by C.CNO;
   

4. 统计每个学生选修课程的门数（超过5门的学生才统计）。要求输出学生学号和选修门数，查询结构按门数降序排列，若门数相同，按学号升序排列

   select SNO, count(distinct CNO)
   from SC
   group by CNO
   having count(*) > 5
   order by count(*) desc, SNO asc;
   

5. 检索学号比Liu同学大， 而年龄比他小的学生的姓名

   select SNAME
   from S as X, S as Y
   where Y.SNAM = 'Liu' and X.SNO > Y.SNO and X.AGE < Y.AGE;
   

6. 在表SC中检索成绩为空值的学生的学号和课程号

   select SNO, CNO
   from SC
   where GRADE is null;
   

7. 检索姓名以L打头的所有学生的姓名和年龄

   select SNAME, AGE
   from S
   where S.name like 'L%';
   

8. 求年龄大于男同学平均年龄的女同学的姓名和年龄

   select SNAME, AGE
   from S
   where S.SEX = '女' and S.AGE > (select avg(Y.AGE)
                                   from S as Y
                                   where Y.SEX = '男');
   

9. 求年龄大于所有女同学年龄的男同学的姓名和年龄

   select SNAME, AGE
   from S
   where S.SEX = '男' and S.AGE > all (select avg(Y.AGE)
                                       from S as Y
                                       where Y.SEX = '女');
   

问题三：试用SQL更新语句表达对数据库中关系S、SC、C的更新操作：

1. 往关系C中插一个课程元组（‘C8’，‘VC++’，‘BAO’）

   insert into C
   	values ('C8', 'VC++', 'BAO');
   

2. 检索所授每门课程平均成绩大于80分的教师姓名，并把检索到的值送往另一个已存在的表FACULTY（TNAME）

   insert into FACULTY(TNAME)
   	select distinct TEACHER
   	from SC as X natural join C
   	where not exists
   		(select CNO
   		 from SC as Y
   		 where X.CNO = Y.CNO and X.SNO = Y.SNO and avg(GRADE) <= 80
   		 group by Y.SNO);
   

3. 在SC中删除尚无成绩的选课元组

   delete from SC
   where SC.GRADE in null;
   

4. 把选修LIU老师课程的女同学选课元组全部删去

   delete from SC
   where SC.SNO in (select distinct SNO
                    from C, S, SC as X
                    where S.SNO = X.SNO and C.CNO = X.CNO and S.SEX = '女' and C.TEACHER = 'LIU');
   

5. 把MATHS课不及格的成绩全改成60分

   update SC
   set SC.GRADE = 60
   where SC.GRADE < 60 and SC.CNO in (select CNO
                                      from C
                                      where C.CNAME = 'MATHS');
   

6. 把低于所有课程总平均成绩的男同学成绩提高5%

   update SC
   set SC.GRADE = SC.GRADE * (1 + 5%)
   where SC.SNO in (select SNO 
                    from S
                    where S.SEX = '男')
   	and SC.GRADE < (select avg(GRADE)
                     	from SC);
   

7. 把表SC中修改C4课程的成绩，若成绩小于等于70分时提高5%，若成绩大于70分时提高4%（用两种方法实现：一种方法是用两个UPDATE语句实现；另一种方法是用带CASE操作的一个UPDATA语句实现）

   -- methods 1:
   update SC
   set SC.GRADE = SC.GRADE * (1 + 5%)
   where SC.GRADE <= 70;
   
   update SC
   set SC.GRADE = SC.GRADE * (1 + 4%)
   where SC.GRADE > 70;
   
   -- methods 2:
   update SC
   set SC.GRADE = case when SC.GRADE <= 70 then SC.GRADE * (1 + 5%)
                       else SC.GRADE * (1 + 4%);
   

8. 在表SC中，当某个成绩低于全部课程的平均成绩时，提高5%

   update SC
   set SC.GRADE = SC.GRADE * (1 + 5%)
   where SC.GRADE < (select avg(GRADE)
                     from SC);
   

问题四：设数据库中有如下三个关系用SQL语句写出下列操作

$EMP(ENO,ENAME,AGE,SEX,ECITY)$ 职工表（职工工号，姓名，年龄，性别，籍贯）

$WORKS(ENO,CNO,SALARY)$ 工作表（职工工号，公司编号，工资）

$COMP(CNO,CNAME,CITY)$公司表（公司编号，公司名称，公司所在城市）

1. 用CREATE TABLE语句创建上述三个表，需指出主键和外键

   -- table EMP
   create table EMP 
   	( ENO	char(15),
         ENAME	varchar(20),
         AGE	numeric(3,0),
         SEX	char(4),
         ECITY	varchar(20),
   	primary key (ENO) );
   
   -- table COMP
   create table COMP
   	( CNO	char(10),
         CNAME	varchar(20),
         CITY	varchar(20),
   	primary key (CNO) );
   
   -- table WORKS
   create table WORKS
   	( ENO		char(15),
         CNO		char(10),
         SALARY	numeric(10,3),
   	primary key (ENO, CNO),
   	foreign key (ENO) references EMP,
       foreign key (CNO) references COMP );
   

2. 检索超过35岁的男职工的工号和姓名

   select ENO, ENAME
   from EMP
   where AGE > 35;
   

3. 假设每个职工只能在一个公司工作，检索工资超过1000元的男性职工的工号和姓名

   select ENO, ENAME
   from EMP, WORKS
   where EMP.ENO = WORKS.ENO and WORKS.SALARY > 1000 and EMP.SEX = '男';
   

4. 假设每个职工可在多个公司工作，检索（同时）在编号为C4和C8公司兼职的职工的工号和姓名

   select distinct ENO, ENAME
   from EMP natural join WORKS
   where WORKS.CNO in (select X.CNO
                       from WORKS as X, WORKS as Y
                       where X.ENO = Y.ENO and X.CNO = 'C4' and Y.CNO = 'C8');
   

5. 检索在 “ 联华公司 ” 工作、工资超过1000元的男性职工的工号和姓名

   select distinct ENO, ENAME
   from EMP natural join WORKS natural join COMP
   where EMP.SEX = '男' and WORKS.SALARY > 1000 and COMP.CNAME = '联华公司';
   

6. 假设每个职工可在多个公司工作，检索每个职工的而兼职公司数目和工资总数，显示$ (ENO,NUM,SUM_SALARY) $分别表示工号、公司数目和工资总数。

   select ENO, count(CNO) as NUM, sum(SALARY) as SUM_SALARY
   from WROKS
   group by ENO;
   

7. 工号为E6的职工在多个公司工作，试检索至少在E6职工兼职的所有公司工作的职工工号

   -- 注意这样的写法会包括E6本人
   select distinct X.ENO
   from WORKS as X
   where not exists (select Y.CNO
                     from WORKS as Y
                     where Y.ENO = 'E6' and not exist (select Z.CNO
                             		   					from WORKS as Z
                                      					where X.ENO = Z.ENO and Z.CNO = Y.CNO) );
   

8. 检索华联公司中低于本公司平均工资的职工的工号和姓名

   select distinct ENO, ENAME
   from WORKS natural join COMP
   where COMP.CNAME = '联华公司' and WORKS.SALARY < 
   	(select avg (SALARY)
        from WORKS natural join COMP
        where COMP.CNAME = '联华公司');
   

9. 在每一公司中为50岁以上职工加薪100元（若职工为多个公司工作，可重复加）

   update WORKS
   set WORKS.SALARY = WORKS.SALARY + 100
   where WORKS.ENO in (select ENO
                       from EMP
                       where EMP.AGE > 50);
   

10. 在EMP表和WORKS表中删除年龄大于60岁的职工有关元组

    -- 注意要先删除WORKS，再删除EMP
    delete from WORKS
    where WORKS.ENO in (select ENO
                        from EMP
                        where EMP.AGE > 60);
    delete from EMP
    where EMP.AGE > 60;
    

问题五：仓库管理数据库中有五个基本表

零件： $PART\ (PNO,PNAME,COLOR,WEIGHT)$

项目： $PROJECT\ (JNO,JNAME,DATE)$

供应商： $SUPPLIER\ (SNO,SNAME,SADDR)$

供应： $P\_P\ (JNO,PNO,TOTAL)$

采购： $P\_S\ (PNO,SNO,QUANTITY)$

1. 试用SQL DDL语句定义上述五个基本表，需说明主键和外键

   -- table PART
   create table PART
   	( PNO		char(12),
         PNAME		varchar(20),
         COLOR		varchar(8),
         WEIGHT	numeric(10,3),
   	primary key (PNO) );
   	
   -- table PROJECT
   create table PROJECT
   	( JNO		char(12),
         JNAME		varchar(20),
         DATE		varchar(100),
       primary key (JNO) );
       
   -- table SUPPLIER
   create table SUPPLIER
   	( SNO		char(12),
         SNAME		varchar(20),
         SADDR		varchar(50),
       primary key (SNO) );
       
   -- table P_P
   create table P_P
   	( JNO		char(12),
         PNO		char(12),
         TOTAL		int
       primary key (JNO, PNO),
       foreign key (JNO) references PROJECT,
       foreign key (PNO) references PART );
       
   -- table P_S
   create table P_S
   	(PNO		char(12),
        SNO		char(12),
        QUANTITY	int,
       primary key (PNO, SNO),
       foreign key (PNO) reference PART,
       foreign key (SNO) reference SUPPLIER );
   

2. 试将PROJECT、P_P、PART三个基本表的连接定义为一个视图VIEW1，将PART、P_S、SUPPLIER三个基本表的连接定义为一个视图VIEW2

   -- create VIEW 1
   create view VIEW1 as
   	select *
   	from PROJECT, P_P, PART
   	where PROJECT.JNO = P_P.JNO and PART.PNO = P_P.PNO;
   
   -- create VIEW 2
   create view VIEW2 as
   	select *
   	from PART, P_S, SUPPLIER
   	where PART.PNO = P_S.PNO and SUPPLIER.SNO = P_S.SNO;
   

3. 试在上述两个视图的基础上进行查询操作

   • 检索上海的供应商所提供的的零件的编号和名称

     select PNO, PNAME
     from VIEW2
     where SEADDR = '上海';
     

   • 检索项目J4所用零件的供应商的编号和名称

     select SNO, SNAME
     from VIEW1
     where JNO = 'J4';
     

问题六：对于问题一的数据库基本表SC，建立如下视图：

create view S_GRADE (SNO, C_NUM, AVG_GRADE) as 
	select SNO, count(CNO), avg(GRADE)
	from SC
	group by SNO;

试判断下列查询和更新操作是否允许执行，如允许，写出转换到基本表SC上的相应操作

1. select * 
   from S_GRADE;
   

   -- 允许操作，等价于：
   select SNO, count(CNO) as C_NUM, avg(GRADE) as AVG_GRADE
   from SC
   group by SNO;
   

2. select SNO, C_NUM 
   from S_GREAD 
   where AVG_GRADE > 80;
   

   -- 允许操作，等价于：
   select SNO, count(CNO) as C_NUM
   from SC
   group by SNO
   having avg(GRADE) > 80;
   

3. select SNO, AVG_GRADE
   from S_GRADE
   where C_NUM > (select C_NUM 
                  from S_GRADE
                  where SNO = 'S4');
   

   -- 允许操作，等价于：
   select X.SNO, avg(GRADE)
   from SC as X
   group by X.SNO;
   having count(X.CNO) > (select count(Y.CNO) 
               		   from SC as Y
                  		   where Y.SNO = 'S4');
   

4. update S_GRADE 
   set SNO = 'S3' 
   where SNO = 'S4';
   

   -- 不允许
   

5. delete from S_GRADE
   where C_NUM > 4
   

   -- 不允许