MySQL经典45题（一）

2022-03-02 19:44:24

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

MySQL经典45题（一）

 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

MySQL经典45题（一）

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‘

MySQL经典45题（一）

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‘

MySQL经典45题（一）

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

MySQL经典45题（一）

 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)

MySQL经典45题（一）

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

MySQL经典45题（一）

4.1 查有成绩的学生信息

select * from
Student
where sid in(SELECT sid from SC)

MySQL经典45题（一）

5.  查询「李」姓老师的数量

SELECT count(1) FROM Teacher
where tname LIKE ‘李%‘

MySQL经典45题（一）

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‘

MySQL经典45题（一）

 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)

MySQL经典45题（一）

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‘)

MySQL经典45题（一）

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‘)

MySQL经典45题（一）

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‘)

MySQL经典45题（一）

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

MySQL经典45题（一）

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

MySQL经典45题（一）

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

MySQL经典45题（一）

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

MySQL经典45题（一）

MySQL经典45题（一）

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

MySQL经典45题（一）

 MySQL经典45题（一）

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

MySQL经典45题（一）

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

MySQL经典45题（一）

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

MySQL经典45题（一）

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

MySQL经典45题（一）

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

MySQL经典45题（一）

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

MySQL经典45题（一）

MySQL经典45题（一）

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