题目1:
====== 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
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
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
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
网上流传的错误答案:
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#] ) )
经过调试验证的正确答案:
SELECT SN, SD FROM
S
WHERE S#
IN ( SELECT
SC.S#
FROM SC RIGHT
JOIN
C
ON SC.C# =
C.C#
GROUP BY
SC.S#
--
在结果集中以学生分组，分组后的 SC.C#选课数=C.C#课程数 即为全部课程
HAVING
COUNT
(
distinct
(SC.C#))
--
注意：一个学生同一门课程可能有多条成绩记录,需要distinct
= (
select
count
(
*
)
from
C )
--
注意：HAVING条件不能用COUNT(distinct(SC.C#)) = COUNT(distinct(C.C#)
)--子查询获得选修全部课程的学生学号
5. 查询选修了课程的学员人数
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
Select SN,SD FROM S
Where [ S# ]
IN
(
Select
[
S#
]
FROM
SC
GROUP BY [
S#
]
HAVING COUNT ( DISTINCT
[
C#
]
)
>
5
)
题目2:
====== 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
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
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
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
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
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
题目3:
======
有如下表记录:
ID Name EmailAddress LastLogon
100 test4 test4@yahoo.cn 2007-11-25 16:31:26
13 test1 test1@yahoo.cn 2007-3-22 16:27:07
19 test1 test1@yahoo.cn 2007-10-25 14:13:46
42 test1 test1@yahoo.cn 2007-11-20 14:20:10
45 test2 test2@yahoo.cn 2007-4-25 14:17:39
49 test2 test2@yahoo.cn 2007-5-25 14:22:36
用一句sql查询出每个用户最近一次登录的记录(每个用户只显示一条最近登录的记录)
方法一:
SELECT a. * from users a inner join
( SELECT [ Name ] , LastLogon = MAX (LastLogon) FROM users
GROUP
BY
[
Name
]
) b
on a. [ Name ] = b. [ Name ]
and
a.
[
LastLogon
]
=
b.
[
LastLogon
]
方法二:
SELECT a. * from users a inner join
( SELECT Name, MAX (LogonID) LogonID FROM users GROUP BY [
Name
]
) b
on a.LogonID = b.LogonID
