MySQL面试经典50题

本文使用的MySQL版本为5.7.21，需要的数据表创建如下：

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

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');

2.科目表 Course(CId,Cname,TId)

--CId --课程编号,Cname 课程名称,TId 教师编号

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');

3.教师表 Teacher(TId,Tname)

--TId 教师编号,Tname 教师姓名

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');

4.成绩表 SC(SId,CId,score)

--SId 学生编号,CId 课程编号,score 分数

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);

利用workbench画出四张表的关系图，表之间通过红框标出来的外键进行关联：

题目

1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

思路：分别查出成绩表中学生“01”和“02”课程的成绩，再使用内连接连接这两个课程的成绩，再利用where进行限制查询
select a.sid,c.Sname,a.score as '01',b.score as '02' from
(select sid,cid,score from SC where cid = '01') as a 
inner join
(select sid,cid,score from SC where cid = '02') as b 
on a.sid = b.sid
inner join student as c on c.sid = a.sid
where a.score>b.score;

运行结果：

1.1 查询同时存在" 01 "课程和" 02 "课程的情况

select a.sid,c.Sname,a.score'01',b.score'02'
from sc as a 
inner join sc as b on a.sid=b.sid
inner join student as c on c.sid = a.sid 
where a.cid='01' and b.cid='02';

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 
(select SId ,score from sc where sc.CId='01')as a 
left join 
(select SId ,score from sc where sc.CId='02')as b
on a.SId=b.SId;

1.3 查询不存在" 01 "课程但存在" 02 "课程的情况

错误写法：这种方法查询出来的结果是不存在“01”且不止存在“02”课程的情况
select * from sc a 
inner join sc b 
on a.sid=b.sid and a.cid='02'
where b.cid not in ('01','02');

结果:

正确写法：
select * from sc a 
where sid not in(select sid from sc where cid='01')
and cid='02';

结果：

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,sname,avg(score) avg_score from sc a
inner join student b on a.sid=b.sid 
group by sid having avg(score)>60;

3.查询在 SC 表存在成绩的学生信息

select b.* from sc a
left join student b
on a.sid=b.sid
group by a.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 a.sid,a.sname,count(b.cid),sum(b.score)
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 a
where a.sid in (select sid from sc group by sid);

5.查询「李」姓老师的数量

select count(tname) from teacher
where tname like '李%';

6.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

7.查询没有学全所有课程的同学的信息

select a.* from student a 
left join sc b 
on a.sid=b.sid 
group by a.sid 
having count(b.cid)<(select count(cid) 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

法一：
select a.* from student a
left join sc b on a.sid=b.sid
where cid in (select cid from sc where sid ='01') and a.sid!='01'
group by a.sid;
法二：
select distinct b.* from sc a 
left join student b 
on a.sid=b.sid 
where cid in (select cid from sc where sid='01') and b.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（重要）

这道题目还是比较棘手，参考了B站up主的方法。

思路：

01号同学学习了01,02,03课程

1.选出所学的课不在（01,02,03）的同学

2.剩下的同学肯定选了01，02，03中的某几门，再判断所学课程数是否等于3

法一：
select * from student
where sid in (
select sid from sc
where sid!='01'
group by sid having count(distinct cid)
=(select count(distinct cid) from sc where sid='01')
)
and sid not in (
select sid from sc where cid not in (
select cid from sc where sid='01')
)

法二;
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 * from student a
where a.sid not in(
select sid from sc b
left join course c
on a.cid=c.cid
inner join teacher d
on c.tid=d.tid and 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,sname,avg(score) as avg_score
from sc a left join student b
on a.sid=b.sid
where a.score<60
group by a.sid having count(cid)>=2;

12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息

select * from sc a 
left join student b on a.sid=b.sid
where cid='01' and score<60
order by 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 c.sid,c.cid,c.score,d.avg_sco
from 
(select a.sid,b.cid,b.score
from student a
left join sc b
on a.sid=b.sid ) c
left join 
 (select sid,avg(a.score) as avg_sco
from sc a
group by a.sid) d
on c.sid=d.sid 
order by avg_sco 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 
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

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

法一：自己没想出来，参考网络其他大佬的解法，自关联，根据比当前分数大的总数来确定排名。

select a.*, count(b.score)+1 as rank
from sc as a 
left join sc as b 
on a.score<b.score and a.cid = b.cid
group by a.cid, a.sid,a.score
order by a.cid, rank;

法二：MySQL8.0及以上版本使用窗口函数进行排名。

sql中的三大排名窗口函数：

三个函数排名效果如下：

本文所使用的版本是MySQL5.7，不支持窗口函数。使用8.0及以上版本：

select *, rank() over(partition by sc.cid order by sc.score desc)排名
from sc;

15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次

法一：可以用变量，不会。

法二：dense_rank函数

select *, dense_rank() over(partition by sc.cid order by sc.score desc)排名
from sc;

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

低版本的MySQL没有提供排名函数真的是很不方便，用其他方法实现三大排名函数的不同排名效果还是没搞明白，下面的解法感觉不是完全符合题意

set @temp=0;
select a.sid, sum, @temp := @temp +1 as rank 
from(
select sc.sid, sum(sc.score) as sum from sc
group by sc.sid
order by sum desc
)a;

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)

成绩分段用case when

select c.cid,c.cname,
sum(case when sc.score<=100 and sc.score>85 then 1 else 0 end) "[100-85)",
sum(case when sc.score<=85 and sc.score>70 then 1 else 0 end) "[85-70)",
sum(case when sc.score<=70 and sc.score>60 then 1 else 0 end) "[70-60)",
sum(case when sc.score<=60 then 1 else 0 end) "[60-0)"
from sc 
inner join course c on sc.cid=c.cid
group by c.cid,c.cname;

18.查询各科成绩前三名的记录

group by后面不能跟limit，换个思路，对于前三名的成绩来说超过他们本身的成绩数量都小于3，如果这个数量等于3或者大于3那么这个成绩肯定就不在前三名之内了

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 cid,count(sid) 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 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 ssex, count(*)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

select * from student a
left join (select sname,ssex,count(*)同名人数 from student group by sname,ssex) b
on a.sname =b.sname and a.ssex=b.ssex
where b.同名人数>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_sco 
from sc 
group by cid 
order by avg_sco 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, a.sname, AVG(sc.score) avg_score from student a, sc
where a.sid = sc.sid
group by sc.sid
having avg_score> 85;

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 a.sname,b.cid,b.score from student a
inner join sc b
on a.sid=b.sid
where cid=(select cid from course where cname='数学')and b.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 * 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 student.sname, course.cname,sc.score from student,course,sc
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

法一：

select distinct sc.CId
from sc
where sc.score <60

法二：

select * from sc a
left join course b
on a.cid=b.cid
where a.score<60 
group by b.cid,b.cname;

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,a.sname
from student a,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,cname,count(sid) 人数 from sc,course c
where sc.cid=c.cid
group by sc.cid;

33.假设成绩不重复，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, sc.cid from student, teacher, course,sc 
where teacher.tid = course.tid
and sc.sid = student.sid
and sc.cid = course.cid
and teacher.tname = "张三"
order by sc.score desc
limit 1;

34.假设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

33题成绩不重复，得到的结果就是唯一的，可以直接limit ，但这一题目，成绩有重复，也就是有可能查出几条同分的最高分，再用limit 就会漏，让score等于最高分的记录就可。

select student.*, sc.score, sc.cid from student, teacher, course,sc 
where teacher.tid = course.tid
and sc.sid = student.sid
and sc.cid = course.cid
and teacher.tname = "张三"
and sc.score = (
    select max(sc.score) 
    from sc,student, teacher, course
    where teacher.tid = course.tid
    and sc.sid = student.sid
    and sc.cid = course.cid
    and teacher.tname = "张三"
);

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

为了避免查询出现重复记录，使用group by

select * from sc a
inner join sc b
on a.sid=b.sid
where a.cid!=b.cid and a.score=b.score
group by a.sid ,a.cid

36.查询每门功成绩最好的前两名

和第18题相同

select * from sc
where (
select count(*) from sc as a 
where sc.cid = a.cid and sc.score<a.score 
)< 2
order by cid asc, sc.score desc;

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 sid, count(cid) as cc from sc
group by sid
having cc>=2;

39.查询选修了全部课程的学生信息

法一;

select * from student a
where (select count(*) from sc b where a.sid=b.sid)
    =(select count(*) from course)

法二：

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 *,year(now())-year(sage) as age
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

timestampdiff 函数不是单纯的计算 年 的相减，月日时间也会算到

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

下面解法有问题，因为每年对应周的日期值并不一定是一样的

select * from student 
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE());

43.查询下周过生日的学生

存在和上题同样的问题

select * from student 
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE())+1;

44.查询本月过生日的学生

select * from student 
where month(student.sage)=month(curdate());

45.查询下月过生日的学生

错误解法：这种解法没有考虑12月的情况，12月的下月是1月，而不是13月

select * from student 
where month(student.sage)=month(curdate())+1;

正解：

select * from student where 
case when month(curdate())=12 then month(student.sage)=1 else month(student.sage)=month(curdate())+1 end;

如有错误，请在评论区指出