数据库常用面试题

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  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　　 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.SNOFROM 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       [e-mail]       2007-11-25 16:31:26
13          test1       [e-mail]       2007-3-22 16:27:07
19          test1       [e-mail]       2007-10-25 14:13:46
42          test1       [e-mail]       2007-11-20 14:20:10
45          test2       [e-mail]       2007-4-25 14:17:39
49          test2       [e-mail]       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
                             --where a.LogonId=b.LogonId

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