【1】-Mysql之超经典SQL练习题

文章下方有视频链接

一、数据表

--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、创建测试数据

1.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');

Student

2.科目表 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');

course

3.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');

teacher

4.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、参考答案

1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

解题思路：此题用的是99连接法(就是9*9乘法公式一样的逻辑)。先找出同时选修了01课程和02课程的学生编号和课程对应的成绩，再用where设置相同条件。

select *
from (select SId ,score from sc where sc.CId='01') as t1 , 
(select SId ,score from sc where sc.CId='02') as t2
where t1.SId=t2.SId
and   t1.score>t2.score

当然也可以用内联查询，我们来试一试：

select * from 
(select sid,cid,score from sc where cid = '01') t1 inner join (select sid,cid,score from sc where cid ='02') t2
on t1.sid=t2.sid and t1.score>t2.score;

1.1 查询同时存在" 01 "课程和" 02 "课程的情况

选修了01课程的学生与选修了02课程的学生的交集即是同时选修了01、02两门课程的学生。

select *
from (select SId ,score from sc where sc.CId='01')as t1 , (select SId ,score from sc where sc.CId='02') as t2
where t1.SId=t2.SId

另外一种表达也可以

select *
from (select SId ,score from sc where sc.CId='01')as t1 inner join (select SId ,score from sc where sc.CId='02') as t2
on t1.SId=t2.SId

再看一种写法：

select *
from (select SId ,score from sc where sc.CId='01')as t1 left  join (select SId ,score from sc where sc.CId='02') as t2
on t1.SId=t2.SId where t2.score is not null;

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 )

思路：此题为典型的左连接题，可以将选修了01课程的学生的成绩信息左连接于选修了02课程的学生的成绩信息。由于采用左连接，所以02课程不存在时自动补充null值。

select *
from (select SId ,score from sc where sc.CId='01')as t1 left join (select SId ,score from sc where sc.CId='02') as t2
on t1.SId=t2.SId

1.3 查询不存在" 01 "课程但存在" 02 "课程的情况

思路：先找出没有选修01课程的学生的成绩信息，再进一步筛选出同时选修了02课程的学生的成绩信息。

select *
from sc
where sc.SId not in (select SId from sc where sc.CId='01')
and  sc.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 sc.sid,st.sname,avg(score) from 
sc,student as st where sc.sid=st.sid group by sc.sid having avg(score)>=60;

来看另外一种思路，有点复杂

select student.*,t1.avgscore
from student inner JOIN(
select sc.SId ,AVG(sc.score)as avgscore
from sc
GROUP BY sc.SId
HAVING AVG(sc.score)>=60)as t1 on student.SId=t1.SId

3.查询在 SC 表存在成绩的学生信息

思路：用sc表中的学生编号对student中的学生信息进行筛选。

select DISTINCT student.*
from student ,sc
where student.SId=sc.SId

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 st.sid,st.sname,count(sc.cid),sum(sc.score) from student st,sc where st.sid=sc.sid group by sc.sid;

另外一种方法

select student.SId,student.Sname,t1.sumscore,t1.coursecount
from student ,(
select SC.SId,sum(sc.score)as sumscore ,count(sc.CId) as coursecount
from sc
GROUP BY sc.SId) as t1
where student.SId =t1.SId

4.1 查有成绩的学生信息

select *
from student
where EXISTS(select * from sc where student.SId=sc.SId)

4.2 查询「李」姓老师的数量

select count(*)
from teacher
where teacher.Tname like '李%

4.3 查询学过「张三」老师授课的同学的信息

select student.*
from teacher  ,course  ,student,sc
where teacher.Tname='张三'
and   teacher.TId=course.TId
and   course.CId=sc.CId
and   sc.SId=student.SId

5. 查询没有学全所有课程的同学的信息

解法1

select student.*
from sc ,student
where sc.SId=student.SId
GROUP BY sc.SId
Having count(*)<(select count(*) from course)

但这种解法得出来的结果不包括什么课都没选的同学。

解法2

select DISTINCT student.*
from
(select student.SId,course.CId
from student,course ) as t1 LEFT JOIN (SELECT sc.SId,sc.CId from sc)as t2 on t1.SId=t2.SId and t1.CId=t2.CId,student
where t2.SId is null
and   t1.SId=student.SId

利用笛卡尔积可以把什么课都没选的同学查询出来

6.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

思路：从sc表中先找出学号为01的学生选修的课程的编号，以选出来的课程编号作为条件，再从sc表中筛选出至少有一门课与学号为01的同学所学相同的同学的学生学号，最后根据筛选出来的学生编号从student表中再次筛选出相应学生的信息。

select DISTINCT student.*
from  sc ,student
where sc.CId in (select CId from sc where sc.SId='01')
and   sc.SId=student.SId

7.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

这道题目也挺难的，这次单独写一篇文章讲解查询和01号学生学习的课程完全相同的其他同学的信息

select t1.* from student t1 inner join
(select sc.sid from sc where cid in 
(select cid from sc where sid ='01') and sid !='01'
group by sc.sid having count(distinct cid) =
(select count(distinct cid) from sc where sid='01')) t2 on t1.sid= t2.sid

8.查询没学过"张三"老师讲授的任一门课程的学生姓名

思路：先查出选了张三将授的课程的学生，再去跟student表进行一个左连接。

 select * from student where student.sid not in
(select distinct student.sid from student,sc where student.sid= sc.sid and  sc.cid = (select course.cid from course,teacher where course.tid=teacher.tid and tname='张三'));

另一种表达：

select * from student where student.SId not in
(select student.SId from student left join sc on student.SId=sc.SId
where EXISTS (select * from teacher ,course where teacher.Tname='张三'
and  teacher.TId=course.TId and   course.CId=sc.CId))

9.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 student.SId,student.Sname,avg(sc.score)
from student ,sc where student.SId=sc.SId and  sc.score<60
GROUP BY sc.SId  HAVING count(*)>=2

10. 检索" 01 "课程分数小于 60，按分数降序排列的学生信息

select student.*
from student,sc
where sc.CId ='01'
and   sc.score<60
and   student.SId=sc.SId

11.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(重点)

select sc.SId,sc.CId,sc.score,t1.avgscore
from  sc left join (select sc.SId,avg(sc.score) as avgscore
from sc
GROUP BY sc.SId) as t1 on sc.SId =t1.SId
ORDER BY t1.avgscore DESC

12.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 sc.CId ,
max(sc.score)as 最高分,min(sc.score)as 最低分,
AVG(sc.score)as 平均分,count(*)as 选修人数,
sum(case when sc.score>=60 then 1 else 0 end )/count(*)as 及格率,
sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end )/count(*)as 中等率,
sum(case when sc.score>=80 and sc.score<90 and sc.score<80 then 1 else 0 end )/count(*)as 优良率,
sum(case when sc.score>=90 then 1 else 0 end )/count(*)as 优秀率
from sc GROUP BY sc.CId
ORDER BY count(*)DESC,sc.CId asc

13.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 sc.CId ,@curRank:=@curRank+1 as rank,sc.score
from (select @curRank:=0) as t ,sc
ORDER BY sc.score desc

14. 按各科成绩进行排序，并显示排名， Score 重复时合并名次

select sc.CId , case when @fontscore=score then @curRank when @fontscore:=score then @curRank:=@curRank+1  end as rank,sc.score
from (select @curRank:=0 ,@fontage:=null) as t ,sc
ORDER BY sc.score desc

15.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 t1.*,@currank:= @currank+1 as rank
from (select sc.SId, sum(score)
from sc
GROUP BY sc.SId
ORDER BY sum(score) desc) as t1,(select @currank:=0) as t

16. 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 t1.*, case when @fontscore=t1.sumscore then @currank  when @fontscore:=t1.sumscore  then @currank:=@currank+1  end as rank
from (select sc.SId, sum(score) as sumscore
from sc
GROUP BY sc.SId
ORDER BY sum(score) desc) as t1,(select @currank:=0,@fontscore:=null) as t

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 course.CId,course.Cname,t1.*
from course LEFT JOIN (
select sc.CId,CONCAT(sum(case when sc.score>=85 and sc.score<=100 then 1 else 0 end )/count(*)*100,'%') as '[85-100]',
CONCAT(sum(case when sc.score>=70 and sc.score<85 then 1 else 0 end )/count(*)*100,'%') as '[70-85)',
CONCAT(sum(case when sc.score>=60 and sc.score<70 then 1 else 0 end )/count(*)*100,'%') as '[60-70)',
CONCAT(sum(case when sc.score>=0 and sc.score<60 then 1 else 0 end )/count(*)*100,'%') as '[0-60)'
from sc
GROUP BY sc.CId) as t1 on course.CId=t1.CId

18.查询各科成绩前三名的记录

思路：前三名转化为若大于此成绩的数量少于3即为前三名。

select *
from sc 
where  (select count(*) from sc as a where sc.CId =a.CId and  sc.score <a.score )<3
ORDER BY CId asc,sc.score desc

19.查询每门课程被选修的学生数


select sc.CId,count(*)
from sc
GROUP BY sc.CId

20.查询出只选修两门课程的学生学号和姓名


select student.SId,student.Sname
from sc,student
where student.SId=sc.SId 
GROUP BY sc.SId
HAVING count(*)=2

21.查询男生、女生人数

select student.Ssex ,count(*) as 人数
from student
GROUP BY student.Ssex

22.查询名字中含有「风」字的学生信息

select *
from student
where student.Sname like '%风%'

23.查询同名同性学生名单，并统计同名人数

select *
from student LEFT JOIN (select Sname,Ssex,COUNT(*)同名人数 from Student group by Sname,Ssex) as t1
on student.Sname =t1.Sname and student.Ssex=t1.Ssex
where t1.同名人数>1

24.查询 1990 年出生的学生名单

select *
from student
where YEAR(student.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 sc.CId,AVG(sc.score)
from sc
GROUP BY sc.CId
ORDER BY AVG(sc.score) desc ,sc.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 student.SId,student.Sname,t1.avgscore
from student INNER JOIN (select sc.SId ,AVG(sc.score) as avgscore from sc GROUP BY sc.SId HAVING AVG(sc.score)>85) as t1 on
student.SId=t1.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 student.Sname ,t1.score
from student INNER JOIN  (select sc.SId,sc.score
from sc,course
where sc.CId=course.CId
and   course.Cname='数学'
and   sc.score<60)as t1 on student.SId=t1.SId

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 student.SId,sc.CId,sc.score from Student  left join sc  on student.SId=sc.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 student.Sname,course.Cname,sc.score
from student , sc  ,course
where sc.score>=70
and  student.SId=sc.SId
and sc.CId=course.CId

30.查询存在不及格的课程

select DISTINCT sc.CId
from sc
where sc.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 student.SId,student.Sname
from student ,sc
where sc.CId='01'
and  student.SId=sc.SId
and  sc.score>80

32.求每门课程的学生人数


select sc.CId,count(*) as 学生人数
from sc
GROUP BY sc.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


select student.*,sc.score
from student ,course ,teacher ,sc
where course.CId=sc.CId
and course.TId=teacher.TId
and teacher.Tname='张三'
and student.SId =sc.SId
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



select student.*,t1.score
from student INNER JOIN (select sc.SId,sc.score, case when @fontage=sc.score then @rank when @fontage:=sc.score then @rank:=@rank+1 end  as rank
from course ,teacher ,sc,(select @fontage:=null,@rank:=0) as t
where course.CId=sc.CId
and course.TId=teacher.TId
and teacher.Tname='张三'
ORDER BY sc.score DESC) as t1 on student.SId=t1.SId
where t1.rank=1

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


select *
from sc as t1
where exists(select * from sc as t2 where t1.SId=t2.SId and t1.CId!=t2.CId and t1.score =t2.score )

36.查询每门功成绩最好的前两名

select *
from sc as t1
where (select count(*) from sc as t2 where t1.CId=t2.CId and t2.score >t1.score)<2
ORDER BY t1.CId

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 sc.CId as 课程编号,count(*) as 选修人数
from sc
GROUP BY sc.CId
HAVING count(*)>5

38.检索至少选修两门课程的学生学号

select DISTINCT t1.SId
from sc as t1
where (select count(* )from sc where t1.SId=sc.SId)>=3

39.查询选修了全部课程的学生信息


select student.*
from sc ,student
where sc.SId=student.SId
GROUP BY sc.SId
HAVING count(*) = (select DISTINCT count(*) from course )

40.查询各学生的年龄，只按年份来算

select student.SId as 学生编号,student.Sname  as  学生姓名,TIMESTAMPDIFF(YEAR,student.Sage,CURDATE()) as 学生年龄
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 student.SId as 学生编号,student.Sname  as  学生姓名,TIMESTAMPDIFF(YEAR,student.Sage,CURDATE()) as 学生年龄
from student

42.查询本周过生日的学生

select *
from student
where YEARWEEK(student.Sage)=YEARWEEK(CURDATE())

43.查询下周过生日的学生


select *
from student
where YEARWEEK(student.Sage)=CONCAT(YEAR(CURDATE()),week(CURDATE())+1)

44.查询本月过生日的学生

select *
from student
where EXTRACT(YEAR_MONTH FROM student.Sage)=EXTRACT(YEAR_MONTH FROM CURDATE())

45.查询下月过生日的学生

select *
from student
where EXTRACT(YEAR_MONTH FROM student.Sage)=EXTRACT(YEAR_MONTH FROM DATE_ADD(CURDATE(),INTERVAL 1 MONTH))

如果想看视频讲解的话
经典50题视频精讲

最后编辑于：2020.03.01 16:20:27

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