设有三个关系：S(S#,SNAME,AGE,SEX)，SC(

(1)

Select C#,CNAME

From C

Where TEACHER=’LIU%’；

(2)

Select S#,SNAME

From S

Where AGE>23 and SEX=’男’；

(3)

Select CNAME,TEACHER

From SC,C

Where SC.C#=C.C# and S# =’S1’;

(4)

Select SNAME

From S,SC,C

Where S.S#=SC.S# and SC.C#=C.C# and TEACHER=’LIU%’and SEX=’女’;

(5)

(Select C#

from C

where C# not in

(select C#

From S,SC

Where S.S#=SC.S# and SNAME=’WANG%’);

(6)

Select S#

From SC

Group by S# having count(*)>=2;

*(7)

[解法1]  select SC.C#,CNAME

From S,SC,C

Where S.S#=SC.S# and SC.C#=C.C#

Group by SC.C#,CNAME having count(*)=

意即：没有一个学生y是不选修课程x。

Select C#, CNAME

From C

Where not exists

( select *

From S

Where not exists

( select *

From SC

Where S#=S.S# and C#=C.C# ));

*(8)

查询学号为x的学生，对任一课程y，只要老师’LIU’讲授了课程y，则学生x也选修了课程y。

假设：若用p表示谓词“老师’LIU’讲授了课程y”，用q表示谓词“学生x选修了课程y”，则上述题意可形式化表示如下：

意即：不存在这样的课程y，老师’LIU’讲授了课程y，而学生x没有选修课程y。

Select S#

From SC SCX

Where not exists

( select *

From C CY

Where TEACHER=’LIU’and not exists

( select *

From SC SCZ

Where CY.C#=SCZ.C# and SCZ.S#=SCX.S# ));