MySQL经典45题

一、数据库字段说明

1、学生表 Student(SId,Sname,Sage,Ssex)
SId ：学生编号
Sname：学生姓名
Sage ：出生年月
Ssex：学生性别

2、课程表 Course(CId,Cname,TId)
CId ：课程编号
Cname ：课程名称
TId ：教师编号

3、教师表 Teacher(TId,Tname)
TId ：教师编号
Tname ：教师姓名

4、成绩表 SC(SId,CId,score)
SId ：学生编号
CId ：课程编号
score：分数

二、插入表及数据

# 学生表 Student：

create table Student(

SId varchar(10) ,

Sname varchar(10),

Sage datetime,

Ssex varchar(10));

insert into Student values('01' , '赵雷' , '1990-01-01' , '男');

insert into Student values('02' , '钱电' , '1990-12-21' , '男');

insert into Student values('03' , '孙风' , '1990-05-20' , '男');

insert into Student values('04' , '李云' , '1990-08-06' , '男');

insert into Student values('05' , '周梅' , '1991-12-01' , '女');

insert into Student values('06' , '吴兰' , '1992-03-01' , '女');

insert into Student values('07' , '郑竹' , '1989-07-01' , '女');

insert into Student values('09' , '张三' , '2017-12-20' , '女');

insert into Student values('10' , '李四' , '2017-12-25' , '女');

insert into Student values('11' , '李四' , '2017-12-30' , '女');

insert into Student values('12' , '赵六' , '2017-01-01' , '女');

insert into Student values('13' , '孙七' , '2018-01-01' , '女');

# 课程表 Course

create table Course(

CId varchar(10),

Cname nvarchar(10),

TId varchar(10)); 

insert into Course values('01' , '语文' , '02'); 

insert into Course values('02' , '数学' , '01'); 

insert into Course values('03' , '英语' , '03');

# 教师表 Teacher

create table Teacher(

TId varchar(10),

Tname varchar(10)); 

insert into Teacher values('01' , '张三');

insert into Teacher values('02' , '李四'); 

insert into Teacher values('03' , '王五');

# 成绩表 SC

create table SC(

SId varchar(10),

CId varchar(10),

score decimal(18,1)); 

insert into SC values('01' , '01' , 80); 

insert into SC values('01' , '02' , 90); 

insert into SC values('01' , '03' , 99); 

insert into SC values('02' , '01' , 70); 

insert into SC values('02' , '02' , 60); 

insert into SC values('02' , '03' , 80); 

insert into SC values('03' , '01' , 80); 

insert into SC values('03' , '02' , 80); 

insert into SC values('03' , '03' , 80); 

insert into SC values('04' , '01' , 50); 

insert into SC values('04' , '02' , 30); 

insert into SC values('04' , '03' , 20); 

insert into SC values('05' , '01' , 76); 

insert into SC values('05' , '02' , 87); 

insert into SC values('06' , '01' , 31); 

insert into SC values('06' , '03' , 34); 

insert into SC values('07' , '02' , 89); 

insert into SC values('07' , '03' , 98);

三、题目与答案

1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

select *

from Student a  INNER JOIN SC b

on a.SId=b.SId

inner join SC c

on a.sid=c.sid AND b.CId=01 and c.CId=02

where b.score > c.score

1.1查询同时存在" 01 "课程和" 02 "课程的情况

SELECT *

FROM

(select * from  SC where cid='01') a inner join

(select * from  SC where cid='02') b on a.sid=b.sid

1.2 查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )

select *

FROM SC a left join SC b

on a.SId=b.SId and b.cid='02'

WHERE a.cid='01'

1.3 查询不存在"01 "课程但存在" 02 "课程的情况

select *

FROM (SELECT * FROM SC WHERE sid not in(SELECT sid FROM SC WHERE cid='01')) a inner join SC b

on a.SId=b.SId and b.cid='02'

2.查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩

SELECT  a.sid,b.sname,a.avg_score

FROM

(SELECT sid,avg(score) as avg_score

FROM SC

GROUP BY sid

HAVING avg_score>=60) a left join Student b

on a.sid=b.sid

3.查询在 SC 表存在成绩的学生信息

select * from

Student

where sid in(SELECT sid from SC)

4. 查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩(没成绩的显示为 null )

select

a.sid,

a.sname,

count(b.cid) as 'scount',

sum(b.score) as 'sum'

from Student a

left join SC b

on a.sid=b.sid

GROUP BY a.sid

4.1 查有成绩的学生信息

select * from

Student

where sid in(SELECT sid from SC)

5. 查询「李」姓老师的数量

SELECT count(1) FROM Teacher

where tname LIKE '李%'

6. 查询学过「张三」老师授课的同学的信息

SELECT a.*

FROM Student a left join SC b on a.sid=b.sid

left join Course c on b.cid=c.cid

left join Teacher d on c.tid=d.tid

WHERE d.tname='张三'

7. 查询没有学全所有课程的同学的信息

SELECT *

FROM Student a left join

SC b on a.sid=b.sid

GROUP BY a.sid

HAVING count(b.cid)<(SELECT COUNT(*) FROM Course)

8. 查询至少有一门课与学号为" 01 "的同学所学相同的同学的信息

SELECT DISTINCT b.*

from SC a inner join Student b on a.sid=b.sid

WHERE a.cid in(SELECT cid from SC where sid='01')

9. 查询和" 01 "号的同学学习的课程完全相同的其他同学的信息

SELECT b.*

FROM

(select * FROM SC where sid not in (SELECT sid FROM SC where cid not in (SELECT cid FROM SC WHERE sid='01') )and sid!='01' )a

left join Student b on a.sid=b.sid

GROUP BY a.sid

HAVING count(cid)=(SELECT count(cid) FROM SC WHERE sid='01')

10. 查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT sid,sname FROM Student WHERE sid not in

(SELECT  DISTINCT a.sid

FROM Student a left join SC b on a.sid=b.sid

left join Course c on b.cid=c.cid

left join Teacher d on c.tid=d.tid

WHERE d.tname='张三')

11. 查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩

select a.sid,

b.sname,

AVG(score) as avg_score

from SC a

left join Student b

on a.sid=b.sid  INNER JOIN (select sid

FROM SC where score<60

GROUP BY sid

having COUNT(1)>1) c on

a.sid=c.sid

group by a.sid

12. 检索" 01 "课程分数小于 60，按分数降序排列的学生信息

SELECT a.* ,b.score

FROM Student  a left join SC b on a.sid=b.sid

WHERE b.cid='01' and b.score<60

ORDER BY b.score desc

13. 按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

SELECT a.*,avg_score

FROM SC a left join

(SELECT sid,avg(score) as avg_score

FROM SC

GROUP BY sid ) b on a.sid=b.sid

ORDER BY avg_score desc

14.查询各科成绩最高分、最低分和平均分：以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90 要求输出课程号和选修人数，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 a.*,b.Cname FROM

(SELECT CId,

MAX(score) as 最高分,

MIN(score) as 最低分,

AVG(score) as 平均分,

COUNT(1)  as 选修人数,

SUM(case when score>=60 then 1 else 0 end) / COUNT(1) as 及格率,

SUM(case when score>=70 and score< 80 then 1 else 0 end) / COUNT(1) as 中等率,

SUM(case when score>=80 and score< 90 then 1 else 0 end) / COUNT(1) as 优良率,

SUM(case when score>=90 then 1 else 0 end) / COUNT(1) as 优秀率

FROM SC

GROUP BY CId) a left join Course b on a.CId=b.CId

ORDER BY 选修人数 DESC,CId ASC

15. 按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

SELECT

sid,cid,score,@rank:=@rank+1 as rk

FROM SC,(SELECT @rank:=0) as t

ORDER BY score desc

15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次

select

*,

case when (@sco=score) then @rank else @rank:=@rank+1 end as rn,

@sco:=score  -- 保存上一次的分数

 from SC ,(select @rank:=0,@sco:=null) as t order by score desc

16. 查询学生的总成绩，并进行排名，总分重复时保留名次空缺

select

s.*,

case when @sco=scos then '' else @rank:=@rank+1 end as rn ,

@sco:=scos

from

(select

sid,sum(score) as scos

from SC group by sid order by scos desc) s,

(select @rank:=0,@sco:=null) as t

16.1 查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

select

a.*,

@RANK:=if(@sco=scos,@rank,@rank+1) as rank,

@sco:=scos

FROM (SELECT sid,sum(score) as scos

FROM SC

GROUP BY sid

ORDER BY scos desc) a,

(SELECT @sco:=null,@rank:=0) b

17. 统计各科成绩各分数段人数：课程编号，课程名称，[100-85]，[85-70]，[70-60]，[60-0] 及所占百分比

select *,

sum(case when 0<=score and score <=60 then 1 else 0 end )/count(1) as '[0,60]',

sum(case when 60<score and score <=70 then 1 else 0 end )/count(1) as '[60,70]',

sum(case when 70<score and score <=85 then 1 else 0 end )/count(1) as '[70,85]',

sum(case when 85<score and score <=100 then 1 else 0 end )/count(1) as '[85,100]'

from SC

group by cid

18. 查询各科成绩前三名的记录

select a.*

from SC a

where (select count(1) from SC b where a.cid=b.cid and b.score>a.score)<3

ORDER BY cid DESC,score DESC

19. 查询每门课程被选修的学生数

SELECT cid,count(1) as cons

from SC

GROUP BY cid

20. 查询出只选修两门课程的学生学号和姓名

SELECT a.sid,b.sname FROM

(select SID,COUNT(1)  AS 选课数量

from SC

GROUP BY SID

HAVING COUNT(1) =2

) a left join Student b on a.sid=b.sid

21. 查询男生、女生人数

SELECT ssex,count(1) as 人数

from Student

GROUP BY ssex

22. 查询名字中含有「风」字的学生信息

SELECT *

FROM Student

WHERE sname like '%风%'

23. 查询同名同性学生名单，并统计同名人数

解题思路：按照姓名分组，姓名形同的情况下按照性别分组统计人数，如果统计人数大于等于1，那说明这个人就是同名同性的

SELECT sname,ssex,count(sname)

FROM Student

GROUP BY sname,ssex

HAVING count(sname)>1

24. 查询 1990 年出生的学生名单

SELECT *

FROM Student

WHERE year(sage)='1990'

25. 查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列

SELECT cid,avg(score) as avg_score

FROM SC

GROUP BY cid

ORDER BY avg_score DESC,cid ASC

26. 查询平均成绩大于等于 85 的所有学生的学号、姓名和平均成绩

SELECT

a.sid,

b.sname,

a.avg_score

FROM

(SELECT sid,avg(score) as avg_score

FROM SC

GROUP BY sid

HAVING avg(score)>=85 ) a left join Student b

on a.sid=b.sid

27. 查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

SELECT b.sname, a.score

FROM SC a

left join Student b on a.sid=b.sid

WHERE a.cid IN

(SELECT cid FROM Course WHERE cname='数学') and a.score<60

28. 查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）

SELECT a.*,b.*

FROM Student a

left join SC b

on a.sid=b.sid

29. 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 b.sname,c.cname,a.score

FROM SC a LEFT JOIN Student b

on a.sid=b.sid LEFT JOIN Course c on a.cid=c.cid

WHERE a.score>70

30. 查询不及格的课程

SELECT distinct cid

FROM SC

WHERE score<60

31. 查询课程编号为 01 且课程成绩在 80 分以上的学生的学号和姓名

SELECT a.sid,b.sname

FROM

(SELECT sid

from SC

where cid='01' and score>=80) a left JOIN Student b on a.sid=b.sid

32. 求每门课程的学生人数

select cid,

count(1) as con

from SC

group by cid

33. 成绩不重复，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

解题思路：4张表连接起来，按成绩排序降序，取第一条记录，就是成绩最高的

SELECT a.*,b.score

FROM Student a  INNER JOIN SC  b on a.sid=b.sid

INNER JOIN Course c on b.cid=c.cid

INNER JOIN Teacher d on c.tid=d.tid

WHERE d.Tname='张三'

ORDER BY b.score desc

LIMIT 1

34. 成绩有重复的情况下，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩（Mysql版本太低，无法使用窗口函数，略）

35. 查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩（Mysql版本太低，无法使用窗口函数，略）

36. 查询每门功成绩最好的前两名（Mysql版本太低，无法使用窗口函数，略）

37. 统计每门课程的学生选修人数（超过 5 人的课程才统计）。

SELECT cid,count(1) as cons

FROM SC

GROUP BY cid

HAVING count(1)>5

38. 检索至少选修两门课程的学生学号

SELECT sid,count(1) as cons

FROM SC

GROUP BY sid

HAVING count(1)>=2

39. 查询选修了全部课程的学生信息

SELECT * from Student where sid IN

(select sid from  SC  GROUP BY sid HAVING COUNT(*)=

(SELECT count(cid) FROM Course) )

40. 查询各学生的年龄，只按年份来算

select *, year(NOW())-year(sage) as 年龄

from Student

41. 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

select *, TIMESTAMPDIFF(YEAR,sage,NOW()) as 年龄

from Student

42. 查询本周过生日的学生

select *,week(Sage),week(now())

from Student

where week(Sage)=week(now());

43. 查询下周过生日的学生

select *,week(Sage),week(now())

from Student

where week(Sage)=week(now())+1;

44. 查询本月过生日的学生

select *,month(Sage),month(now())

from Student

where month(Sage)=month(now());

45. 查询下月过生日的学生

select *,month(Sage),month(now())

from Student

where month(Sage)=month(now())+1;

巴特西

MySQL经典45题

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