2024mysql面试题

本人最近在自学sql，从开始学到自己写完本练习50题大概花了12天的时间。
学习路径：《sql基础教程》第1遍（3天）→知乎中的sql网课+leetcode刷题（4天）→牛客网刷题（2天）→《sql基础教程》第2遍（1晚上）→sql面试练习50题（3天）
文中的代码都是按自己思路写的，大家在做题的时候最好也按照自己的思路来写，实在想不通的再参考我的思路，毕竟每个人的编写思路不一样，唯有自己思考过才会印象深刻。有问题的小伙伴可以评论留言，一起探讨~

表名和字段

Student(s_id,s_name,s_birth,s_sex) --学生编号,学生姓名, 出生年月,学生性别

–2.课程表

Course(c_id,c_name,t_id) – --课程编号, 课程名称, 教师编号

–3.教师表

Teacher(t_id,t_name) --教师编号,教师姓名

–4.成绩表

Score(s_id,c_id,s_score) --学生编号,课程编号,分数

测试数据

--建表
--学生表
CREATE TABLE `Student`(
	`s_id` VARCHAR(20),
	`s_name` VARCHAR(20) NOT NULL DEFAULT '',
	`s_birth` VARCHAR(20) NOT NULL DEFAULT '',
	`s_sex` VARCHAR(10) NOT NULL DEFAULT '',
	PRIMARY KEY(`s_id`)
);
--课程表
CREATE TABLE `Course`(
	`c_id`  VARCHAR(20),
	`c_name` VARCHAR(20) NOT NULL DEFAULT '',
	`t_id` VARCHAR(20) NOT NULL,
	PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
	`t_id` VARCHAR(20),
	`t_name` VARCHAR(20) NOT NULL DEFAULT '',
	PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
	`s_id` VARCHAR(20),
	`c_id`  VARCHAR(20),
	`s_score` INT(3),
	PRIMARY KEY(`s_id`,`c_id`)
);
--插入学生表测试数据
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('08' , '王菊' , '1990-01-20' , '女');
--课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');

--教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

--成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);

练习题目及sql语句

1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数

-- [分析] 筛选出课程号01这门课的成绩比02这门课成绩高的学生，输出这些学生的信息及课程分（表：Student，Score)——表联结
select a.*, b.s_score as 01_score, c.s_score as 02_score
from Student as a inner join Score as b on a.s_id = b.s_id and b.c_id = '01'
inner join Score as c on a.s_id = c.s_id and c.c_id = '02' 
where b.s_score > c.s_score;

2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数

-- [分析] 与上体的思路一样，换个符合方向就可
select a.*, b.s_score as 01_score, c.s_score as 02_score
from student as a inner join score as b on a.s_id = b.s_id and b.c_id = '01'
inner join score as c on a.s_id = c.s_id and c.c_id = '02'
where b.s_score < c.s_score;

3、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

-- [分析] 要求平均成绩，就需要用group by对学生分组，然后利用聚合函数avg求出平均成绩，由于where语句中不能包含聚合函数，故再利用having语句和60分比较。
-- 输出的结果学生编号和学生姓名在student表中，成绩在score表中，故需要用到内联结。
select a.s_id, a.s_name, avg(s_score) as avg_score
from student as a inner join score as b on a.s_id = b.s_id
group by a.s_id,a.s_name
having avg(b.s_score) >= 60;

4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩

-- [分析] 和3题思路一样，改个符号方向即可
select a.s_id, a.s_name, avg(s_score) as avg_score
from student as a inner join score as b on a.s_id = b.s_id
group by b.s_id
having avg(b.s_score) < 60;

5、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

-- [分析]学生编号、学生姓名在表student中；选课数可通过在表score中通过group by 学生编号后利用count(c_id)算出；所有课程的总成绩则用sum可以算出。
-- 因为要用到表student和表score，故需要对表进行联结。
select a.s_id, a.s_name, count(b.c_id), sum(b.s_score)
from student as a inner join score as b on a.s_id = b.s_id
group by a.s_id,a.s_name;

6、查询"李"姓老师的数量

-- [分析] 这里用到知识点字符串模糊查询，如：where 姓名 like '猴%' 即可查到猴什么什么。
-- 关于字符串模糊查询的知识点可查看我的另一篇文章《SQL学习笔记——汇总分析》
select count(t_name) as number
from teacher
where t_name like '李%';

7、查询学过"张三"老师授课的同学的信息

-- [分析] 老师的信息在表teacher中，可通过teacher.t_id与表course联结，
-- 再通过course.c_id与表score联结，
-- 再再通过score.s_id与表student联结，查找出上过张三老师课的同学的信息。
select s.*
from teacher as a inner join course as b on a.t_id = b.t_id
inner join score as c on b.c_id = c.c_id
inner join student as s on c.s_id = s.s_id
where a.t_name = '张三';

8、查询没学过"张三"老师授课的同学的信息

-- [分析]这里不能直接将“=”替换成“<>”，因为不等于张三时，还有别的老师。所以要用到子查询来解决。
select *
from student 
where s_id not in 
(select s.s_id
from teacher as a inner join course as b on a.t_id = b.t_id
inner join score as c on b.c_id = c.c_id
inner join student as s on c.s_id = s.s_id
where a.t_name = '张三');

9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息

-- [分析]由于既要学过01课程的，又要学过02课程的，所以共需要联结两次才能达到筛选效果
select *
from student
where s_id in (select a.s_id from student as a inner join score as b on a.s_id = b.s_id and b.c_id = '01' 
                      inner join score as c on b.s_id = c.s_id and c.c_id = '02');

10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息

-- [分析] 在上述语句中用and联结，注意不能直接将“=”换成“<>”
select *
from student 
where s_id in (select a.s_id from student as a inner join score as b on a.s_id = b.s_id and b.c_id = '01')
and s_id not in (select a.s_id from student as a inner join score as c on a.s_id = c.s_id and c.c_id = '02');

11、查询没有学全所有课程的同学的信息

-- [分析]没学全所有课程的情况很多，如学0门，学1门，学2门。所以这里先筛选出学全所有课程的同学再取反，即没有学全所有课程的同学。
select *
from student
where s_id not in (select s_id 
                   from score 
                   group by s_id 
                   having count(c_id) = (select count(distinct c_id) from course));

12、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 *

-- [分析]子查询两次，第一次找到学号为01的同学所学的课程，第二次再找到至少一门课与学号01同学上的课相同的同学信息
select *
from student 
where s_id in (select distinct a.s_id from score as a 
                       where a.c_id in(select b.c_id from score as b 
                                                 where b.s_id = '01')) and s_id <> '01';

-- 方法2：内联结+一次子查询
select distinct a.*
from student as a inner join score as b on a.s_id = b.s_id 
where b.c_id in (select c_id from score where s_id = '01') 
and b.s_id <> '01' ;

13、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 （不会）

-- [分析] 利用子查询 查出学号01的课程数 再去找课程数一样的其他同学  
select student.*
from student
where s_id in (select s_id from score group by s_id having count(c_id) = (select count(c_id) from score where s_id = '01')
and s_id <> '01');

14、查询没学过"张三"老师讲授的任一门课程的学生姓名

-- [分析] 方法一：疯狂子查询

select s_name
from student
where s_id not in (select s_id from score 
                                 where c_id = (select c_id from course
                                                           where t_id = (select t_id from teacher 
                                                                                     where t_name = '张三')));

方法二：子查询+内联结

select s_name
from student
where s_id not in (select a.s_id from score as a inner join course as b on a.c_id = b.c_id
                   inner join teacher as c on b.t_id = c.t_id and c.t_name = '张三');

15、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

-- [分析] 内联结+筛选
select a.s_id,a.s_name,avg(b.s_score) as avg_score
from student as a inner join score as b on a.s_id = b.s_id
where a.s_id in (select s_id from score where s_score < 60 group by s_id having count(s_score) >= 2)
group by a.s_id,a.s_name; -- where筛选出来后一定不要忘记分组显示！！！
-- 方法2 （简单）
select a.s_id,a.s_name,avg(b.s_score)
from student as a inner join score as b on a.s_id = b.s_id and b.s_score < 60
group by a.s_id,a.s_name
having count(b.s_score) >= 2;

16、检索"01"课程分数小于60，按分数降序排列的学生信息

-- [分析] 内联结+子查询筛选+排序
select a.*,b.c_id,b.s_score
from student as a inner join score as b on a.s_id = b.s_id and b.c_id = 01
where a.s_id in (select s_id from score where s_score < 60  order by s_score desc);

17、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（重要！）

select a.s_id,(select s_score from score where s_id = a.s_id and c_id = '01') as 语文,
              (select s_score from score where s_id = a.s_id and c_id = '02') as 数学,
              (select s_score from score where s_id = a.s_id and c_id = '03') as 英语,
              round(avg(s_score),2) as 平均分 
from score as a 
group by a.s_id 
order by 平均分 desc;

-- 18.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（重要！）

-- [分析] 用到case语句，后面记得按课程号分组来显示全部结果
select a.c_id,a.c_name,max(b.s_score),min(b.s_score),avg(b.s_score),
       round(100*(sum(case when b.s_score >= 60 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end)),2) as 及格率,
       round(100*(sum(case when b.s_score between 70 and 80 then 1 else 0 end)/sum(case when b.s_score then 1 end)),2) as 中等率,
		   round(100*(sum(case when b.s_score between 80 and 90 then 1 else 0 end)/sum(case when b.s_score then 1 end)),2) as 优良率,
       round(100*(sum(case when b.s_score >= 90 then 1 else 0 end)/sum(case when b.s_score then 1 end)),2) as 优秀率
from course as a inner join score as b on a.c_id = b.c_id
group by a.c_id,a.c_name;

19、按各科成绩进行排序，并显示排名（涉及到TopN问题，重要！！！）

-- [分析] 本题涉及到窗口函数的使用 重点记住语句：dense_rank() over (partition by 列名 order by 排序用列 desc
select s_id, c_id, s_score, dense_rank() over (PARTITION by c_id order by s_score desc) as rank_
from score;

20、查询学生的总成绩并进行排名

select s_id,sum(s_score)
from score
group by s_id
order by sum(s_score) desc;

21、查询不同老师所教不同课程平均分从高到低显示

select a.t_id,avg(b.s_score)
from course as a inner join score as b on a.c_id = b.c_id
group by a.t_id
order by avg(b.s_score) desc;

22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩

select a.*,sum(b.s_score)
from student as a inner join score as b on a.s_id = b.s_id
group by s_id -- 一定不要忘记分组啊啊啊啊啊！
order by sum(b.s_score) desc
limit 1,2 -- 1代表把排名第一的数去了，2代表去除第一名后往下筛选两个出来

23、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],[0-60]及所占百分比

select a.c_id, a.c_name,
sum(case when b.s_score between 85 and 100 then 1 else 0 end) as '100-85',
round(100*(sum(case when b.s_score between 85 and 100 then 1 end)/sum(case when b.s_score then 1 end)),2),
sum(case when b.s_score between 70 and 85 then 1 end) as '85-70',
round(100*(sum(case when b.s_score between 75 and 85 then 1 end)/sum(case when b.s_score then 1 end)),2),
sum(case when b.s_score between 60 and 70 then 1 end) as '70-60',
round(100*(sum(case when b.s_score between 60 and 70 then 1 end)/sum(case when b.s_score then 1 end)),2),
sum(case when b.s_score < 60 then 1 end) as 不及格,
round(100*(sum(case when b.s_score <=60 then 1 end)/sum(case when b.s_score then 1 end)),2)
from course as a inner join score as b on a.c_id = b.c_id
group by a.c_id,a.c_name

24、查询学生平均成绩及其名次

select s_id,avg(s_score),rank() over (order by avg(s_score) desc) as rank_score
from score
group by s_id

25、查询各科成绩前三名的记录（重要！！！！！）

-- [分析]这道题最简单的写法就是分别写出课程号，然后用union关键字合并
(select * from score where c_id = '01' order by s_score desc limit 3)
UNION
(select * from score where c_id = '02' order by s_score desc limit 3)
UNION
(select * from score where c_id = '03' order by s_score desc limit 3);

26、查询每门课程被选修的学生数

-- [分析] 有点简单，写完甚至有点难以置信，还亲自去数了一遍。。。。。。
select c_id, count(s_id)
from score
group by c_id;

27、查询出只有两门课程的全部学生的学号和姓名

-- [分析] 内联结+筛选
select a.s_id,a.s_name
from student as a inner join score as b on a.s_id = b.s_id
group by a.s_id
having count(b.c_id) = 2;

28、查询男生、女生人数

方法1：
select s_sex,count(s_sex) as 人数
from student
group by s_sex;
方法2：
(select s_sex,count(s_sex) as 人数 from student where s_sex = '男')
union
(select s_sex,count(s_sex) as 人数 from student where s_sex = '女')

29、查询名字中含有"风"字的学生信息

-- [分析] 考察字符串模糊查询，之前我的sql学习笔记中有详细记录https://zhuanlan.zhihu.com/p/128093194

select *
from student
where s_name like "%风%";

30、查询同名同性学生名单，并统计同名人数（重要！！！！！）

-- [分析] 将表student定义两次，然后进行内联结，并给出判断条件
select a.s_name,a.s_sex,count(*)
from student as a inner join student as b on a.s_id <> b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex
group by a.s_name,a.s_sex;

31、查询1990年出生的学生名单

-- [分析] 字符串模糊查询
select *
from student
where s_birth like "1990%";

32、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

-- [分析] 考察order by 语句的升降序及先后次序
select c_id,avg(s_score)
from score
group by c_id
order by avg(s_score) desc,c_id asc;

33、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.s_id, a.s_name, avg(b.s_score)
from student as a inner join score as b on a.s_id = b.s_id
group by b.a_id,a.s_name
having avg(b.s_score) >= 85;

34、查询课程名称为"数学"，且分数低于60的学生姓名和分数

-- [分析] 三个表联结，两个筛选条件：①c.cname = '数学' ②b.s_score < 60
select a.s_name, b.s_score
from student as a inner join score as b on a.s_id = b.s_id
inner join course as c on b.c_id = c.c_id and c.c_name = '数学'
where b.s_score < 60;

35、查询所有学生的课程及分数情况；

select a.s_id, b.c_id, b.c_name, a.s_score
from score as a inner join course as b on a.c_id = b.c_id;

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 a.s_name,c.c_name,b.s_score
from student as a inner join score as b on a.s_id = b.s_id
inner join course as c on c.c_id = b.c_id
where b.s_score >= 70;

37、查询不及格的课程

select a.s_id,a.s_name,c.c_id,c.c_name,b.s_score
from student as a inner join score as b on a.s_id = b.s_id
inner join course as c on b.c_id = c.c_id
where b.s_score < 60;

38、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.s_id,a.s_name
from student as a inner join score as b on a.s_id = b.s_id
where b.c_id = '01' and b.s_score >= 80;

39、求每门课程的学生人数

select a.c_id,a.c_name,count(b.s_id)
from course as a inner join score as b on a.c_id = b.c_id
group by b.c_id;

40、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 d.*,max(c.s_score)
from teacher as a inner join course as b on a.t_id = b.t_id
inner join score as c on c.c_id = b.c_id
inner join student as d on d.s_id = c.s_id
where a.t_name = '张三'

41、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 （重要！！！！）

select distinct b.s_id,b.c_id,b.s_score 
from score as a,score as b 
where a.c_id != b.c_id and a.s_score = b.s_score

42、查询每门功成绩最好的前两名

-- [分析]这道题不吹牛逼，我认为我的写法最牛逼
(select c_id,s_score from score where c_id = '01' order by s_score desc limit 2)
UNION
(select c_id,s_score from score where c_id = '02' order by s_score desc limit 2)
UNION
(select c_id,s_score from score where c_id = '03' order by s_score desc limit 2)

43、统计每门课程的学生选修人数（超过5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，若人数相同，按课程号升序排列

select c_id,count(s_id)
from score
group by c_id
having count(s_id) >= 5
order by count(s_id) desc,c_id asc;

44、检索至少选修两门课程的学生学号

select s_id,count(c_id)
from score
group by s_id
having count(c_id) >= 2;

45、查询选修了全部课程的学生信息

select a.*
from student as a inner join score as b on a.s_id = b.s_id
group by s_id
having count(b.c_id) = (select count(c_id) from course);

46、查询各学生的年龄 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

-- [分析] 这道题俺不会，学sql 的时候没学日期函数的使用，答案是查来的
select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') - 
(case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age
from student;

47、查询本周过生日的学生

select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

48、查询下周过生日的学生

select*from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1=WEEK(s_birth)

49、查询本月过生日的学生

select*from student whereMONTH(DATE_FORMAT(NOW(),'%Y%m%d'))=MONTH(s_birth)

50、查询下月过生日的学生

select*from student whereMONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1=MONTH(s_birth)