Student(S#,Sname,Sage,Ssex) 学生表
Course(C#,Cname,T#) 课程表
SC(S#,C#,score) 成绩表
Teacher(T#,Tname) 教师表
问:查询没有学全所有课的同学的学号、姓名
下面是我的,我总感觉应该还有一条更好的select st.sno,sname
from student st ,sc
where st.sno=sc.sno
group by sno
having count(sc.cno) <> ( select count(1) from course)
Course(C#,Cname,T#) 课程表
SC(S#,C#,score) 成绩表
Teacher(T#,Tname) 教师表
问:查询没有学全所有课的同学的学号、姓名
下面是我的,我总感觉应该还有一条更好的select st.sno,sname
from student st ,sc
where st.sno=sc.sno
group by sno
having count(sc.cno) <> ( select count(1) from course)
from student st ,sc
where st.sno=sc.sno
group by sno
--个人觉得加个distinct比较好
having count(distinct sc.cno) <> ( select count(1) from course)
select x1.s#,x1.name from(
select x.s# learned from 学生表 x inner join 成绩表 c on x.s#=c.s# j
inner join 课程表 k on k.c#=c.c# group by s# having count(*)>(select count(1)
from 课程表 group by c# )t inner join 学生表 x1 x1.s#=x.s#
--好像有比你的还麻烦!
select st.sno,sname
from student st
where st.sno in (select sc.sno from sc group by sno
having count(sc.cno) <> ( select count(1) from course)) 这样子是不是更好点
(
select * from student ,course
) a
where not exists(select 1 from sc b where a.s#=b.s# and a.c#=b.c#)
SQL codeselect st.sno,sname
from student st
where st.sno in (select sc.sno from sc group by sno
having count(sc.cno) <> ( select count(1) from course))
这样子是不是更好点
---
问题描述:
为管理岗位业务培训信息,建立3个表:
S (S#,SN,SD,SA) S#,SN,SD,SA 分别代表学号、学员姓名、所属单位、学员年龄
C (C#,CN ) C#,CN 分别代表课程编号、课程名称
SC ( S#,C#,G ) S#,C#,G 分别代表学号、所选修的课程编号、学习成绩
1. 使用标准SQL嵌套语句查询选修课程名称为’税收基础’的学员学号和姓名
--实现代码:
SELECT SN,SD FROM S
WHERE [S#] IN(
SELECT [S#] FROM C,SC
WHERE C.[C#]=SC.[C#]
AND CN=N'税收基础')
2. 使用标准SQL嵌套语句查询选修课程编号为’C2’的学员姓名和所属单位
--实现代码:
SELECT S.SN,S.SD FROM S,SC
WHERE S.[S#]=SC.[S#]
AND SC.[C#]='C2'
3. 使用标准SQL嵌套语句查询不选修课程编号为’C5’的学员姓名和所属单位
--实现代码:
SELECT SN,SD FROM S
WHERE [S#] NOT IN(
SELECT [S#] FROM SC
WHERE [C#]='C5')
4. 使用标准SQL嵌套语句查询选修全部课程的学员姓名和所属单位
--实现代码:
SELECT SN,SD FROM S
WHERE [S#] IN(
SELECT [S#] FROM SC
RIGHT JOIN
C ON SC. [C#] =C. [C#] GROUP BY [S#]
HAVING COUNT (*) =COUNT ([S#]))
5. 查询选修了课程的学员人数
--实现代码:
SELECT 学员人数=COUNT(DISTINCT [S#]) FROM SC 6. 查询选修课程超过5门的学员学号和所属单位
--实现代码:
SELECT SN, SD FROM S
WHERE [S#] IN (
SELECT [S#] FROM SC
GROUP BY [S#]
HAVING COUNT (DISTINCT [C#])>5)
题目2
问题描述:
已知关系模式:
S (SNO,SNAME) 学生关系。SNO 为学号,SNAME 为姓名
C (CNO,CNAME,CTEACHER) 课程关系。CNO 为课程号,CNAME 为课程名,CTEACHER 为任课教师
SC(SNO,CNO,SCGRADE) 选课关系。SCGRADE 为成绩
1. 找出没有选修过“李明”老师讲授课程的所有学生姓名
--实现代码:
SELECT SNAME FROM S
WHERE NOT EXISTS (
SELECT * FROM SC, C
WHERE SC.CNO=C.CNO
AND CNAME='李明'
AND SC.SNO=S.SNO)
2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩
--实现代码:
SELECT S.SNO, S.SNAME, AVG_SCGRADE=AVG (SC.SCGRADE)
FROM S, SC, (
SELECT SNO
FROM SC
WHERE SCGRADE<60
GROUP BY SNO
HAVING COUNT(DISTINCT CNO)>=2
)A WHERE S.SNO=A.SNO AND SC.SNO=A.SNO
GROUP BY S.SNO,S.SNAME 3. 列出既学过“1”号课程,又学过“2”号课程的所有学生姓名
--实现代码:
SELECT S.SNO,S.SNAME
FROM S,(
SELECT SC.SNO
FROM SC,C
WHERE SC.CNO=C.CNO
AND C.CNAME IN('1','2')
GROUP BY SNO
HAVING COUNT (DISTINCT CNO) =2
)SC WHERE S.SNO=SC.SNO
4. 列出“1”号课成绩比“2”号同学该门课成绩高的所有学生的学号
--实现代码:
SELECT S.SNO, S.SNAME
FROM S, (
SELECT SC1.SNO
FROM SC SC1, C C1, SC SC2, C C2
WHERE SC1.CNO=C1.CNO AND C1.NAME='1'
AND SC2.CNO=C2.CNO AND C2.NAME='2'
AND SC1.SCGRADE>SC2.SCGRADE
)SC WHERE S.SNO=SC.SNO 5. 列出“1”号课成绩比“2”号课成绩高的所有学生的学号及其“1”号课和“2”号课的成绩
--实现代码:
SELECT S.SNO,S.SNAME,SC.[1号课成绩],SC.[2号课成绩]
FROM S,(
SELECT SC1.SNO,[1号课成绩]=SC1.SCGRADE,[2号课成绩]=SC2.SCGRADE
FROM SC SC1,C C1,SC SC2,C C2
WHERE SC1.CNO=C1.CNO AND C1.NAME='1'
AND SC2.CNO=C2.CNO AND C2.NAME='2'
AND SC1.SCGRADE>SC2.SCGRADE
)SC WHERE S.SNO=SC.SNO