MySQL查询语句练习题 50题版

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学生表:Student(编号sid,姓名sname,年龄sage,性别ssex)
课程表:Course(课程编号cid,课程名称cname,教师编号tid)
成绩表:Sc(学生编号sid,课程编号cid,成绩score)
教师表:Teacher(教师编号tid,姓名tname)

1,插入学生数据

学号姓名年龄性别
1001张三10
1002李四11
1003王五12
1004马六19
1005孙七22
1006钱八18
1007赵九11
1008周公19

2,插入老师数据

教师编号教师姓名
1叶平
2李浩然
3胡平原
4朱清时
5赛先生
6宋三东

3,插入课程数据

课程编号课程名教师编号
001PHP1
002C#1
003C++2
004JAVA3
005Python4
006R5
007HTML6

4,插入成绩数据

学号课程编号成绩
100100189
100200180
100300130
100400178
100500168
100600193
100700162
100100267
100200286
100300267
100400277
100500266
100600284
100700272
100100382
100200385
100300332
100400373
100500364
100600387
100700377
100800394
100100439
100200480
100300480
100400488
100500468
100600459
100700442
100800464
100100589
100200570
100300560
100400558
100500538
100600589
100700572
100800564
100100649
100200690
100300670
100400648
100500698
100600659
100700672
100800674
100100749
100200750
100300770
100400788
100500778
100600799
100700782

以下题目答案均由45题版本的数据库来操作。懒得再建了。

1、查询“001”课程比“002”课程成绩高的所有学生的学号;

#自连接
select Sno from Score a 
where Cno= '3-105' and 
Degree > (select Degree from Score b where Cno='3-245'and a.Sno=b.Sno);
#输出两个表后连接比较
select a.Sno from 
(select Sno,Degree from Score where Cno='3-105') a inner join 
(select Sno,Degree from Score where Cno='3-245') b 
on a.Sno=b.Sno
where a.Degree>b.Degree;

2、查询平均成绩大于60分的同学的学号和平均成绩;

select Sno,avg(Degree) from Score 
group by Sno
having avg(Degree)>60;

3、查询所有同学的学号、姓名、选课数、总成绩;

select Score.Sno,Sname,count(*),sum(Degree) from Score,Student where Score.Sno=Student.Sno group by Score.Sno ;

select Score.Sno,Sname,count(*),sum(Degree) from Score inner join Student on Score.Sno=Student.Sno group by Sno;

4、查询姓“李”的老师的个数;

#还需要去重
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 Sno,Sname from Student where Sno not in (
select Sno from Score where Cno in (
select Cno from Course where Tno = (
select Tno from Teacher where Tname ='李诚')));

select Sno,Sname from Student where Sno not in(
select Score.Sno from Score inner join Course on Score.Cno=Course.Cno inner join Teacher on Course.Tno=Teacher.Tno where Tname='李诚');

6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;

select Student.Sno,Sname from Student inner join Score on Student.Sno=Score.Sno where Cno='001' and Student.Sno in (select Student.Sno,Sname from Student inner join Score on Student.Sno=Score.Sno where Cno='002');

select Sno,Sname from Student where Sno in(
select Sno from Score where Cno = '001' and Sno in(
select Sno from Score where Cno='002'));

select Student.Sno,Student.Sname from Student,Score  where Student.Sno=Score.Sno and Score.Sno='001' and exists( Select * from Score as SC_2 where SC_2.Sno=SC.Sno and SC_2.Sno='002');
#EXISTS用于检查子查询是否至少会返回一行数据,该子查询实际上并不返回任何数据,而是返回值True或False。

7、查询学过“叶平”老师所教的所有课的同学的学号、姓名;

select Sno,Sname from Student where Sno in (
select Sno from Score where Cno in(
select Cno from Course where Tno =(
select Tno from Teacher where Tname = '李诚'))
group by Sno 
having count(*)=(select count(*) from Course,Teacher where Course.Tno=Teacher.Tno and Tname ='李诚')
);

select Student.Sno,Student.Sname from Student,Score,Course,Teacher where Student.Sno=Score.Sno
and Score.Cno=Course.Cno and Course.Tno = Teacher.Tno and Teacher.Tname='李诚'
group by Sno 
having count(*) = (select count(*) from Course,Teacher where Course.Tno= Teacher.Tno and Teacher.Tname= '李诚');

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 Sno,Sname from Student where Sno in (
select Sno from Score a2 where Cno='002' and Degree>
(select Degree from Score a1 where Cno='001' and a1.Sno=a2.Sno));

Select Sno,Sname from (select Student.Sno,Student.Sname,Degree,(select score from Score SC_2 where SC_2.Sno=Student.Sno and SC_2.Cno='002') Degree2 from Student,Score where Student.Sno=Score.Sno and Cno ='001') S_2 where Degree2 <Degree;

9、查询所有课程成绩小于60分的同学的学号、姓名;

#注意关键词所有课程,可以采用逆向思维
select Sno,Sname from Student where Sno not in (
select Sno from Score where Degree>60);

10、查询没有学全所有课的同学的学号、姓名;

select Sno,Sname from Student where Sno not in (
select Sno from Score group by Sno
having count(*) = (
select count(distinct(Cno)) from Score));

select Sno,Sname from Student where Sno not in (
select Sno from Score group by Sno
having count(*) = (
select count(*) from Course));

select Student.Sno,Student.Sname
from Student,Score
where Student.Sno=Score.Sno group by  Student.Sno,Student.Sname 
having count(Cno) <(select count(Cno) from Course); 

11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名;

select Sno,Sname from Student where Sno in (
select Sno from Score where Cno in (
select Cno from Score where Sno = '103'));

12、查询与“1001”同学所学课程相同的其他同学学号和姓名;

#自己做的待考证
select Sno,Sname from Student where Sno in(
select Sno from Score a where Cno in (select Cno from Score where Sno='103')
group by Sno 
having count(*) = (select count(*) from Score where Sno='103')
);

13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩;

#这个方法出不来
update Score a set Degree = (select avg(Degree) from Score b group by Cno where a.Cno=b.Cno)
where a.Cno in (select Cno from Course,Teacher where Teacher.Tno = Course.Tno and Tname='李诚'));
#网上的方法是
update SC set score=(select avg(SC_2.score) 
    from SC SC_2 
    where SC_2.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='叶平'); 

update Score inner join (select Cno,avg(Degree) avgDegree from Score where Cno in 
(select Cno from Course where Tno = (select Tno from Teacher where Tname = '李诚')) group by Cno) a
on Score.Cno=a.Cno
Set Score.degree = a.avgDegree;

14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名;

#有可能学了同样数目但是类型不同的课程,所以用not in
select Sno,Sname from Student where Sno in (
select Sno from Score where Sno not in(
select Sno from Score where Cno not in (select Cno from Score where Sno = '103'))
group by Sno
having count(Cno) = (select count(Cno) from Score where Sno = '103'));

15、删除学习“叶平”老师课的SC表记录;

delete from Score where Cno = (
select Cno from Course where Tno = (
select Tno from Teacher where Tname='叶平'));

16、向SC表中插入一些记录,这些记录要求符合以下条件:没有上过编号“003”课程的同学学号、’002’号课的平均成绩;

insert into Score (Sno,Cno,Degree)
values
(select Sno,002,(select avg(Degree) from Score where Cno ='002' ) from Student where Sno not in (select Sno from Score where Cno = '003'));

17、按平均成绩从高到低显示所有学生的“JAVA”、“C#”、“C++”三门的课程成绩,按如下形式显示: 学生ID,,C#,C++,JAVA,有效课程数,有效平均分。

select Sno as '学生ID',
(select Degree from Score a where Cno =(select Cno from Course where Cname='C#') and a.Sno=SC.Sno) as 'C#',
(select Degree from Score b where Cno =(select Cno from Course where Cname='C++') and b.Sno=SC.Sno) as 'C++',
(select Degree from Score c where Cno =(select Cno from Course where Cname='JAVA') and c.Sno=SC.Sno) as 'JAVA',
count(*) as '有效课程数',
avg(Degree) as '有效平均分'
from Score SC group by Sno 
order by avg(Degree) desc;

18、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分;

#自己写的
select Cno as '课程ID',max(Degree) as '最高分',min(Degree) as '最低分' from Score group by Cno;
#网上的
SELECT L.Cno As 课程ID,L.score AS 最高分,R.score AS 最低分 
    FROM SC L ,SC AS R 
    WHERE L.Cno = R.Cno and 
        L.Degree = (SELECT MAX(IL.Degree) 
                      FROM SC AS IL,Student AS IM 
                      WHERE L.Cno = IL.Cno and IM.Sno=IL.Sno 
                      GROUP BY IL.Cno) 
        AND 
        R.Score = (SELECT MIN(IR.Degree) 
                      FROM SC AS IR 
                      WHERE R.Cno = IR.Cno 
                  GROUP BY IR.Cno
                    ); 

19、按各科平均成绩从低到高和及格率的百分数从高到低顺序 ;

#自己写的
select * from Score a order by 
(select avg(Degree) from Score b where a.Cno=b.Cno group by Cno) asc,
(select count(*) from Score c where Degree > 60 and a.Cno=c.Cno group by Cno)/(select count(*) from Score d where a.Cno=d.Cno group by Cno) desc;
#网上的
SELECT t.C# AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS平均成绩,100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数
    FROM SC T,Course
    where t.C#=course.C#
    GROUP BY t.C#
    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 C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分,
100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数,
SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分,
100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数,
SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分,
100* SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数,
SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分,
100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '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.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# AS 课程ID,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.C#=C.C# and C.T#=Z.T# 
  GROUP BY C.C# 
  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.S# 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.S# = T1.S# AND T1.C# = '001' 
            LEFT JOIN SC AS T2 
                      ON SC.S# = T2.S# AND T2.C# = '002' 
            LEFT JOIN SC AS T3 
                      ON SC.S# = T3.S# AND T3.C# = '003' 
            LEFT JOIN SC AS T4 
                      ON SC.S# = T4.S# AND T4.C# = '004' 
      WHERE student.S#=SC.S# 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) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) 
      FROM sc 
            LEFT JOIN sc AS T1 
                      ON sc.S# = T1.S# AND T1.C# = 'k1' 
            LEFT JOIN sc AS T2 
                      ON sc.S# = T2.S# AND T2.C# = 'k2' 
            LEFT JOIN sc AS T3 
                      ON sc.S# = T3.S# AND T3.C# = 'k3' 
            LEFT JOIN sc AS T4 
                      ON sc.S# = T4.S# AND T4.C# = 'k4' 
      ORDER BY ISNULL(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.C# 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) AS [85 - 70] 
        ,SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 - 60] 
        ,SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS [60 -] 
    FROM SC,Course 
    where SC.C#=Course.C# 
    GROUP BY SC.C#,Cname;

24、查询学生平均成绩及其名次

      SELECT 1+(SELECT COUNT( distinct 平均成绩) 
              FROM (SELECT S#,AVG(score) AS 平均成绩 
                      FROM SC 
                  GROUP BY S# 
                  ) AS T1 
            WHERE 平均成绩 > T2.平均成绩) as 名次, 
      S# as 学生学号,平均成绩 
    FROM (SELECT S#,AVG(score) 平均成绩 
            FROM SC 
        GROUP BY S# 
        ) 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.S# as 学生ID,t1.C# as 课程ID,Score as 分数 
      FROM SC t1 
      WHERE score IN (SELECT TOP 3 score 
              FROM SC 
              WHERE t1.C#= C# 
            ORDER BY score DESC 
              ) 
      ORDER BY t1.C#; 

26、查询每门课程被选修的学生数

select c#,count(S#) from sc group by C#; 

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.S#,Student.Sname,count(C#) AS 选课数 
  from SC ,Student 
  where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=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 Student 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 C#,Avg(score) from SC group by C# order by Avg(score),C# 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.S# ,avg(score) 
    from Student,SC 
    where Student.S#=SC.S# group by SC.S#,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.S#=Student.S# and SC.C#=Course.C# and  Course.Cname='数据库'and score <60; 

35、查询所有学生的选课情况;

    SELECT SC.S#,SC.C#,Sname,Cname 
    FROM SC,Student,Course 
    where SC.S#=Student.S# and SC.C#=Course.C# ; 

36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数;

    SELECT  distinct student.S#,student.Sname,SC.C#,SC.score 
    FROM student,Sc 
    WHERE SC.score>=70 AND SC.S#=student.S#; 

37、查询不及格的课程,并按课程号从大到小排列

select c# from sc where scor e <60 order by C# ; 

38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名;

select SC.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='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.S#=SC.S# and SC.C#=C.C# and C.T#=Teacher.T# and Teacher.Tname='叶平' and SC.score=(select max(score)from SC where C#=C.C# ); 

41、查询各个课程及相应的选修人数

select count(*) from sc group by C#; 

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.S#,B.score from SC A  ,SC B where A.Score=B.Score and A.C# <>B.C# ; 

43、查询每门功成绩最好的前两名

SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数 
      FROM SC t1 
      WHERE score IN (SELECT TOP 2 score 
              FROM SC 
              WHERE t1.C#= C# 
            ORDER BY score DESC 
              ) 
      ORDER BY t1.C#; 

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  C# as 课程号,count(*) as 人数 
    from  sc  
    group  by  C# 
    order  by  count(*) desc,c#  

45、检索至少选修两门课程的学生学号

select  S#  
    from  sc  
    group  by  s# 
    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  C#,Cname  
    from  Course  
    where  C#  in  (select  c#  from  sc group  by  c#)

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 S# not in (select S# from Course,Teacher,SC where Course.T#=Teacher.T# and SC.C#=course.C# 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 S#,avg(isnull(score,0)) from SC where S# in (select S# from SC where score <60 group by S# having count(*)>2)group by S#;

49、检索“004”课程分数小于60,按分数降序排列的同学学号

select S# from SC where C#='004'and score <60 order by score desc; 

50、删除“002”同学的“001”课程的成绩

delete from Sc where S#='001'and C#='001'; 

其他练习题:
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

要求实现如下5个处理:

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

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(DISTINCT [S#]))

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)

点这里可以跳转到人工智能网站

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