经典SQL面试题.doc

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1、如有侵权，请联系网站删除，仅供学习与交流经典SQL面试题【精品文档】第 14 页表Student(Sid,Sname,Sage,Ssex) 学生表 CREATE TABLE student ( sid varchar(10) NOT NULL, sName varchar(20) DEFAULT NULL, sAge datetime DEFAULT 1980-10-12 23:12:36, sSex varchar(10) DEFAULT NULL, PRIMARY KEY (sid) ENGINE=InnoDB DEFAULT CHARSET=utf8;Course(Cid,Cname,T

2、id) 课程表 CREATE TABLE course ( cid varchar(10) NOT NULL, cName varchar(10) DEFAULT NULL, tid int(20) DEFAULT NULL, PRIMARY KEY (cid) ENGINE=InnoDB DEFAULT CHARSET=utf8;SC(Sid,Cid,score) 成绩表 CREATE TABLE sc ( sid varchar(10) DEFAULT NULL, cid varchar(10) DEFAULT NULL, score int(10) DEFAULT NULL) ENGIN

3、E=InnoDB DEFAULT CHARSET=utf8;Teacher(Tid,Tname) 教师表 CREATE TABLE taacher ( tid int(10) DEFAULT NULL, tName varchar(10) DEFAULT NULL) ENGINE=InnoDB DEFAULT CHARSET=utf8;数据：（MySQL）insert into taacher(tid,tName) values (1,李老师),(2,何以琛),(3,叶平);insert into student(sid,sName,sAge,sSex) values (1001,张三丰,19

4、80-10-12 23:12:36,男),(1002,张无极,1995-10-12 23:12:36,男),(1003,李奎,1992-10-12 23:12:36,女),(1004,李元宝,1980-10-12 23:12:36,女),(1005,李世明,1981-10-12 23:12:36,男),(1006,赵六,1986-10-12 23:12:36,男),(1007,田七,1981-10-12 23:12:36,女);insert into sc(sid,cid,score) values (1,001,80),(1,002,60),(1,003,75),(2,001,85),(2,

5、002,70),(3,004,100),(3,001,90),(3,002,55),(4,002,65),(4,003,60);insert into course(cid,cName,tid) values (001,企业管理,3),(002,马克思,3),(003,UML,2),(004,数据库,1),(005,英语,1);问题： 1、查询“001”课程比“002”课程成绩高的所有学生的学号; select a.Sid from (select sid,score from SC where Cid=001) a,(select sid,score from SC where Cid=00

6、2) b where a.scoreb.score and a.sid=b.sid; 2、查询平均成绩大于60分的同学的学号和平均成绩; select Sid,avg(score) from sc group by Sid having avg(score) 60; 3、查询所有同学的学号、姓名、选课数、总成绩; select Student.Sid,Student.Sname,count(SC.Cid),sum(score) from Student left Outer join SC on Student.Sid=SC.Sid group by Student.Sid,Sname 4、查

7、询姓“李”的老师的个数; select count(distinct(Tname) from Teacher where Tname like 李%; 5、查询没学过“叶平”老师课的同学的学号、姓名; select Student.Sid,Student.Sname from Student where Sid not in (select distinct( SC.Sid) from SC,Course,Teacher where SC.Cid=Course.Cid and Teacher.Tid=Course.Tid and Teacher.Tname=叶平); 6、查询学过“001”并且

8、也学过编号“002”课程的同学的学号、姓名; select Student.Sid,Student.Sname from Student,SC where Student.Sid=SC.Sid and SC.Cid=001and exists( Select * from SC as SC_2 where SC_2.Sid=SC.Sid and SC_2.Cid=002); 7、查询学过“叶平”老师所教的所有课的同学的学号、姓名; select Sid,Sname from Student where Sid in (select Sid from SC ,Course ,Teacher wh

9、ere SC.Cid=Course.Cid and Teacher.Tid=Course.Tid and Teacher.Tname=叶平 group by Sid having count(SC.Cid)=(select count(Cid) from Course,Teacher where Teacher.Tid=Course.Tid and Tname=叶平); 8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名; Select Sid,Sname from (select Student.Sid,Student.Sname,score ,(select

10、score from SC SC_2 where SC_2.Sid=Student.Sid and SC_2.Cid=002) score2 from Student,SC where Student.Sid=SC.Sid and Cid=001) S_2 where score2 60); 10、查询没有学全所有课的同学的学号、姓名; select Student.Sid,Student.Sname from Student,SC where Student.Sid=SC.Sid group by Student.Sid,Student.Sname having count(Cid) =60

11、 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数 FROM SC T,Course where t.Cid=course.Cid GROUP BY t.Cid ORDER BY 100 * SUM(CASE WHEN isnull(score,0)=60 THEN 1 ELSE 0 END)/COUNT(*) DESC 20、查询如下课程平均成绩和及格率的百分数(用1行显示): 企业管理（001），马克思（002），OO&UML （003），数据库（004） SELECT SUM(CASE WHEN Cid =001 THEN score ELSE 0 END)/SU

12、M(CASE Cid WHEN 001 THEN 1 ELSE 0 END) AS 企业管理平均分 ,100 * SUM(CASE WHEN Cid = 001 AND score = 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cid = 001 THEN 1 ELSE 0 END) AS 企业管理及格百分数 ,SUM(CASE WHEN Cid = 002 THEN score ELSE 0 END)/SUM(CASE Cid WHEN 002 THEN 1 ELSE 0 END) AS 马克思平均分 ,100 * SUM(CASE WHEN Cid = 002

13、 AND score = 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cid = 002 THEN 1 ELSE 0 END) AS 马克思及格百分数 ,SUM(CASE WHEN Cid = 003 THEN score ELSE 0 END)/SUM(CASE Cid WHEN 003 THEN 1 ELSE 0 END) AS UML平均分 ,100 * SUM(CASE WHEN Cid = 003 AND score = 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cid = 003 THEN 1 ELSE 0 END) AS

14、UML及格百分数 ,SUM(CASE WHEN Cid = 004 THEN score ELSE 0 END)/SUM(CASE Cid WHEN 004 THEN 1 ELSE 0 END) AS 数据库平均分 ,100 * SUM(CASE WHEN Cid = 004 AND score = 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cid = 004 THEN 1 ELSE 0 END) AS 数据库及格百分数 FROM SC 21、查询不同老师所教不同课程平均分从高到低显示 SELECT max(Z.Tid) AS 教师ID,MAX(Z.Tname)

15、AS 教师姓名,C.Cid AS 课程,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩 FROM SC AS T,Course AS C ,Teacher AS Z where T.Cid=C.Cid and C.Tid=Z.Tid GROUP BY C.Cid ORDER BY AVG(Score) DESC 22、查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理（001），马克思（002），UML （003），数据库（004） 学生ID,学生姓名,企业管理,马克思,UML,数据库,平均成绩 SELECT DISTINCT top 3 SC.Sid

16、 As 学生学号, Student.Sname AS 学生姓名 , T1.score AS 企业管理, T2.score AS 马克思, T3.score AS UML, T4.score AS 数据库, ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分 FROM Student,SC LEFT JOIN SC AS T1 ON SC.Sid = T1.Sid AND T1.Cid = 001 LEFT JOIN SC AS T2 ON SC.Sid = T2.Sid

17、AND T2.Cid = 002 LEFT JOIN SC AS T3 ON SC.Sid = T3.Sid AND T3.Cid = 003 LEFT JOIN SC AS T4 ON SC.Sid = T4.Sid AND T4.Cid = 004 WHERE student.Sid=SC.Sid and ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) NOT IN (SELECT DISTINCT TOP 15 WITH TIES ISNULL(T1.score,0) +

18、ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) FROM sc LEFT JOIN sc AS T1 ON sc.Sid = T1.Sid AND T1.Cid = k1 LEFT JOIN sc AS T2 ON sc.Sid = T2.Sid AND T2.Cid = k2 LEFT JOIN sc AS T3 ON sc.Sid = T3.Sid AND T3.Cid = k3 LEFT JOIN sc AS T4 ON sc.Sid = T4.Sid AND T4.Cid = k4 ORDER BY ISNULL

19、(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC); 23、统计列印各科成绩,各分数段人数:课程ID,课程名称,100-85,85-70,70-60, 60 SELECT SC.Cid as 课程ID, Cname as 课程名称 ,SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS 100 - 85 ,SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END)

20、 AS 85 - 70 ,SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS 70 - 60 ,SUM(CASE WHEN score T2.平均成绩) as 名次, Sid as 学生学号,平均成绩 FROM (SELECT Sid,AVG(score) 平均成绩 FROM SC GROUP BY Sid ) AS T2 ORDER BY 平均成绩 desc; 25、查询各科成绩前三名的记录:(不考虑成绩并列情况) SELECT t1.Sid as 学生ID,t1.Cid as 课程ID,Score as 分数 FROM S

21、C t1 WHERE score IN (SELECT TOP 3 score FROM SC WHERE t1.Cid= Cid ORDER BY score DESC ) ORDER BY t1.Cid; 26、查询每门课程被选修的学生数 select cid,count(Sid) from sc group by Cid; 27、查询出只选修了一门课程的全部学生的学号和姓名 select SC.Sid,Student.Sname,count(Cid) AS 选课数 from SC ,Student where SC.Sid=Student.Sid group by SC.Sid ,Stu

22、dent.Sname having count(Cid)=1; 28、查询男生、女生人数 Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex=男; Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex=女; 29、查询姓“张”的学生名单 SELECT Sname FROM Student WHERE Sname like 张%; 30、查询同名同性学生名单，并统计同名人数 select Sname,count(*) from Studen

23、t group by Sname having count(*)1; 31、1981年出生的学生名单(注：Student表中Sage列的类型是datetime) select Sname, CONVERT(char (11),DATEPART(year,Sage) as age from student where CONVERT(char(11),DATEPART(year,Sage)=1981; 32、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列 Select Cid,Avg(score) from SC group by Cid order by Av

24、g(score),Cid DESC ; 33、查询平均成绩大于85的所有学生的学号、姓名和平均成绩 select Sname,SC.Sid ,avg(score) from Student,SC where Student.Sid=SC.Sid group by SC.Sid,Sname having avg(score)85; 34、查询课程名称为“数据库”，且分数低于60的学生姓名和分数 Select Sname,isnull(score,0) from Student,SC,Course where SC.Sid=Student.Sid and SC.Cid=Course.Cid and

25、 Course.Cname=数据库and score =70 AND SC.Sid=student.Sid; 37、查询不及格的课程，并按课程号从大到小排列 select cid from sc where scor e 80 and Cid=003; 39、求选了课程的学生人数 select count(*) from sc; 40、查询选修“叶平”老师所授课程的学生中，成绩最高的学生姓名及其成绩 select Student.Sname,score from Student,SC,Course C,Teacher where Student.Sid=SC.Sid and SC.Cid=C.

26、Cid and C.Tid=Teacher.Tid and Teacher.Tname=叶平 and SC.score=(select max(score)from SC where Cid=C.Cid ); 41、查询各个课程及相应的选修人数 select count(*) from sc group by Cid; 42、查询不同课程成绩相同的学生的学号、课程号、学生成绩 select distinct A.Sid,B.score from SC A ,SC B where A.Score=B.Score and A.Cid B.Cid ; 43、查询每门功成绩最好的前两名 SELECT

27、t1.Sid as 学生ID,t1.Cid as 课程ID,Score as 分数 FROM SC t1 WHERE score IN (SELECT TOP 2 score FROM SC WHERE t1.Cid= Cid ORDER BY score DESC ) ORDER BY t1.Cid; 44、统计每门课程的学生选修人数（超过10人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，查询结果按人数降序排列，若人数相同，按课程号升序排列 select Cid as 课程号,count(*) as 人数 from sc group by Cid order by co

28、unt(*) desc,cid 45、检索至少选修两门课程的学生学号 select Sid from sc group by sid having count(*) = 2 46、查询全部学生都选修的课程的课程号和课程名 select Cid,Cname from Course where Cid in (select cid from sc group by cid) 47、查询没学过“叶平”老师讲授的任一门课程的学生姓名 select Sname from Student where Sid not in (select Sid from Course,Teacher,SC where C

29、ourse.Tid=Teacher.Tid and SC.Cid=course.Cid and Tname=叶平); 48、查询两门以上不及格课程的同学的学号及其平均成绩 select Sid,avg(isnull(score,0) from SC where Sid in (select Sid from SC where score 2)group by Sid; 49、检索“004”课程分数小于60，按分数降序排列的同学学号 select Sid from SC where Cid=004and score 60 order by score desc; 50、删除“002”同学的“001”课程的成绩 delete from Sc where Sid=001and Cid=001;