深耕MySQL - 50道SQL练习题

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文章目录

1. 数据准备

1、有以下4张表：

学生表：student(学号,学生姓名,出生年月,性别)

成绩表：score(学号,课程号,成绩)

课程表：course(课程号,课程名称,教师号)

教师表：teacher(教师号,教师姓名)

2、创建表结构并插入数据：

drop table if exists Student ;

-- 创建学生表
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);

3、数据库表：



2. 50道SQL面试题

1、查询课程编号为“01”的课程比“02”的课程成绩高的所有学生的学号（难）

select distinct a.s_id 
from ( select * from score where c_id=01) a 
join ( select * from score where c_id=02) b 
on a.s_score>b.s_score 
where a.s_id=b.s_id;

思路：

(1) 先求出课程编号为“01”的学生课程成绩

(2) 再求出课程编号为“02”的学生课程成绩

(3) 将这两张表连接查询，条件是a.s_id = b.s_id，代表是同一个学生

2、查询平均成绩大于60分的学生的学号和平均成绩

(1) 求每个学生的平均成绩
(2) 使用having对每个学生的平均成绩过滤

3、查询所有学生的学号、姓名、选课数、总成绩

select a.s_name,b.s_id,b.count,b.sum_score 
from student a 
left join (select s_id,count(*) count,sum(s_score) sum_score 
           from score 
           group by s_id) b 
on a.s_id=b.s_id;

思路：

(1) 先查询出所有学生的学号，选课数，总成绩

(2) 将步骤1中查询出来的这张表作为一个字表与student表进行连接查询

简化查询：可以不使用子查询，直接将两张表连接查询

select a.s_name,b.s_id,count(b.s_id) count,sum(b.s_score) sum_score 
from student a  
left join score b 
on a.s_id=b.s_id 
group by s_id;

4、查询姓“猴”的老师的个数

mysql> select count(t_name) from teacher where t_name='猴';

5、查询没学过“张三”老师课的学生的学号、姓名

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

思路：

(1) 首先，将4张表连接，查询出选过张三老师课的学生的学号

select a.s_id
from score a 
join student b on a.s_id=b.s_id
join course c on a.c_id=c.c_id
join teacher d on c.t_id=d.t_id
where d.t_name='张三';


(2) 将步骤1中的查询作为一个子查询，与student表精进行连接查询

6、查询学过“张三”老师所教的所有课的同学的学号、姓名

select b.s_id,b.s_name 
from score a 
join student b on a.s_id=b.s_id 
join course c on a.c_id=c.c_id 
join teacher d on c.t_id=d.t_id 
where d.t_name='张三';

7、查询学过编号为“01”的课程并且也学过编号为“02”的课程的学生的学号、姓名

方法1 ：

select s_id,s_name 
from student 
where s_id in(select s_id 
              from score 
              where c_id='01')  
and s_id 
           in(select s_id 
              from score 
              where c_id='02');

思路：
(1) 分别查询学过编号为“01” 和“02”的课程的学生的学号

select s_id from score where c_id='01'
select s_id from score where c_id='02'


(2) 学过编号为“01”的课程并且也学过编号为“02”的课程的学生的学号、姓名

方法2 ：

select distinct s.s_id,s.s_name from score a
join (select s_id from score where c_id='01') b on a.s_id=b.s_id
join (select s_id from score where c_id='02') c on a.s_id=c.s_id
join student s on a.s_id=s.s_id;

思路：

(1) 先将三张表连接查询学过编号为“01”的课程并且也学过编号为“02”的课程的学生的学号

select distinct a.s_id 
from score a 
join (select s_id from score where c_id='01') b on a.s_id=b.s_id 
join (select s_id from score where c_id='02') c on a.s_id=c.s_id;


(2) 将student表和步骤1的查询结果连表查询出sname

8、查询课程编号为“02”的总成绩

mysql> select sum(s_score) from score group by c_id having c_id='02';

9、查询所有，课程成绩小于60分的学生的学号、姓名

select distinct a.s_id,a.s_name 
from student a 
left join score b on a.s_id=b.s_id 
where b.s_score<60 or a.s_id not in (select s_id from score);

(1) 查询课程成绩小于60分的学生的学号

(2) 将步骤1的查询结果和student表联表查询出姓名，同时考虑没有考试成绩的学生

10、查询没有学全所有课的学生的学号、姓名

select a.s_id,a.s_name 
from student a 
left join score b on a.s_id=b.s_id 
group by a.s_id,a.s_name
having count(b.c_id)<(select count(c_id) from course);

11、查询至少有一门课与学号为“01”的学生所学课程相同的学生的学号和姓名 （难）

select distinct a.s_id,a.s_name 
from student a 
right join score b on a.s_id=b.s_id 
where b.c_id in (select c_id from score where s_id='01') and a.s_id !='01';

(1) 查询至少有一门课与学号为“01”的学生所学课程相同的学生的学号

(2) 根据步骤1的结果与student表联表查询出姓名，同时排除学号1的学生

12、查询和“01”号同学所学课程完全相同的其他同学的学号（难）

select s_id 
from score where s_id!='01' 
group by s_id 
having count(s_id) = (select count(*) from score where s_id='01');

13、查询没学过"张三"老师讲授的任一门课程的学生姓名

select s_id,s_name 
from student 
where s_id not in (select a.s_id 
                   from score a 
                   join course b on a.c_id=b.c_id 
                   join teacher c on b.t_id=c.t_id 
                   where 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

select a.s_id,a.s_name,avg(s_score) avg_score 
from student a 
join score b on a.s_id=b.s_id 
where b.s_score<60 
group by s_id 
having count(b.s_id)>=2;

(1) 求出每个学生成绩不及格的课程数量及平均成绩

(2) 利用步骤1的结果与student表联表查询出学生姓名，其中课程成绩不及格数量要大于等于2门

16、检索"01"课程分数小于60，按分数降序排列的学生信息

select a.s_id,a.s_name 
from student a 
join score b on a.s_id=b.s_id 
where b.s_score<60 and b.c_id='01' 
order by b.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(难)

select a.s_id,a.s_score,b.avg_score
from score a 
left join (select s_id, avg(s_score) avg_score 
           from score 
           group by s_id) b 
on a.s_id =b.s_id 
order by avg_score desc;


注意：上述虽然表达出了题目的意思，但是最好以横排形式展示，可以使用select嵌套子查询

方法2 ：

select s_id,s_name,
(select s_score from score where score.s_id=student.s_id and c_id='01') 01_score,
(select s_score from score where score.s_id=student.s_id and c_id='02') 02_score,
(select s_score from score where score.s_id=student.s_id and c_id='03') 03_score,
(select avg(s_score) from score where score.s_id=student.s_id) avg_score
from student 
order by avg_score desc;

使用select子查询嵌套查询时，需要将有歧义的字段区分出来比如： score.s_id=student.s_id

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

(1) 查询各科成绩最高分、最低分和平均分

select a.c_id,a.c_name, avg(b.s_score) avg_score,min(b.s_score) min_score,max(b.s_score) max_score 
from course a j
oin score b on a.c_id=b.c_id 
group by b.c_id;


(2) 每一个科目的及格率，中等率，优良率，优秀率需要使用select后嵌套子查询获得

select 
b.c_id,b.c_name, avg(a.s_score) avg_score,min(a.s_score) min_score,max(a.s_score) max_score,
-- 注意：这里相当于将 score a和course b联表查询后再与score这张表联表查询
(select count(*) from score where c_id=a.c_id and s_score>=60)/count(*) '及格率',
(select count(*) from score where c_id=a.c_id and s_score between 70 and 80)/count(*) '中登率',
(select count(*) from score where c_id=a.c_id and s_score between 80 and 90)/count(*) '优良率',
(select count(*) from score where c_id=a.c_id and s_score>90) /count(*) '优秀率'
from score a
join course b
on a.c_id=b.c_id
group by a.c_id;

19、查询学生的总成绩并进行排名

select a.s_id,a.s_name,(case when sum(b.s_score) is null then 0 else sum(b.s_score) end) sum_score 
from student a 
left join score b on a.s_id=b.s_id 
group by b.s_id 
order by sum_score desc;

20、查询不同老师所教不同课程平均分,从高到低显示

select c_id,avg(s_score) avg_score 
from score 
group by c_id 
order by avg_score desc;

21、查询学生平均成绩及其名次

select s_id,avg(s_score) avg_score,
row_number () over( order by avg(s_score) desc) row_num
from score 
group by s_id;

22、按各科成绩进行排序，并显示排名(难)

序号函数：row_number() / rank() / dense_rank()

partition子句：窗口按照那些字段进行分组，窗口函数在不同的分组上分别执行。下面的例子就按照 c_id进行了分组。在每个 c_id上，按照order by的顺序分别生成从1开始的顺序编号。

order by子句：按照哪些字段进行排序，窗口函数将按照排序后的记录顺序进行编号。可以和partition子句配合使用，也可以单独使用。如果没有partition子句，则会按照所有score排序来生成序号。

-- partition by c_id 按照c_id进行分组
--  order by s_score desc 按照s_score倒叙排序
select c_id,s_id,row_number() over(partition by c_id order by s_score desc) num_row 
from score;

23、查询每门功课成绩最好的前两名学生姓名

select a.s_id,a.s_name,b.num_row 
from student a 
-- 将select查询作为一张子表，与student表联表查询
join (select 
      s_id,
      c_id,
      row_number() over(partition by c_id order by s_score desc) num_row
      from score) b on a.s_id=b.s_id 
where b.num_row<=2;

24、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.s_id,a.s_name,b.c_id,b.s_score,b.num_row 
from student a 
join (select 
      s_id,
      c_id,
      s_score,
      row_number() over(partition by c_id order by s_score desc) num_row from score) b 
on a.s_id=b.s_id 
where b.num_row between 2 and 3;

25、查询各科成绩前三名的记录（不考虑成绩并列情况）

select a.s_id,a.s_name,b.c_id,b.s_score,b.num_row 
from student a 
join (select 
      s_id,
      c_id,
      s_score,row_number() over(partition by c_id order by s_score desc) num_row 
      from score) b 
on a.s_id=b.s_id 
where b.num_row<=3;

26、使用分段[100-85],[85-70],[70-60],[<60]来统计各科成绩，分别统计各分数段人数：课程ID和课程名称

select a.c_id,a.c_name,
(select count(*) from score where score.c_id=a.c_id and s_score<60) '小于60',
(select count(*) from score where score.c_id=a.c_id and s_score between 60 and 70) '60-70',
(select count(*) from score where score.c_id=a.c_id and s_score between 70 and 85) '70-85',
(select count(*) from score where score.c_id=a.c_id and s_score between 85 and 100) '85-100'
from course a 
join score b on a.c_id=b.c_id 
group by c_id;

27、查询每门课程被选修的学生数

mysql> select c_id,count(c_id) count from score group by c_id;

28、查询出只有两门课程的全部学生的学号和姓名

select a.s_id,a.s_name,count(b.s_id) count 
from student a 
join score b on a.s_id=b.s_id 
group by s_id 
having count=2;

29、查询男生、女生人数

mysql> select count(s_sex) count_sex from student group by s_sex;

30、查询名字中含有"风"字的学生信息

mysql> select * from student where s_name like '%周%';

31、查询1990年出生的学生名单

mysql> select s_id,s_name from student where year(s_birth)=1990;

32、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) avg_score 
from student a 
join score b on a.s_id=b.s_id 
group by b.s_id 
having avg_score>=85;

33、查询每门课程的平均成绩，结果按平均成绩升序排序，平均成绩相同时，按课程号降序排列

select c_id, avg(s_score) avg_score from score group by c_id order by avg_score,c_id desc;

34、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.c_id,b.c_name,c.s_id,c.s_name,a.s_score
from score a 
join course b on a.c_id=b.c_id 
join student c on a.s_id=c.s_id 
where a.s_score<60 and b.c_name='数学';

35、查询所有学生的课程及分数情况

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

37、查询不及格的课程并按课程号从大到小排列

mysql> select distinct c_id from score where s_score<60 order by c_id desc;

38、查询课程编号为03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 a 
join score b on a.s_id=b.s_id 
where b.s_score>80 and b.c_id=03;

39、求每门课程的学生人数

mysql> select c_id, count(*) count from score group by 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姓名及其成绩

select a.s_score,b.s_id,b.s_name 
from score a 
join student b on a.s_id=b.s_id 
join course c on a.c_id=c.c_id 
join teacher d on c.t_id=d.t_id 
where d.t_name='张三' 
order by a.s_score desc 
limit 1;

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 （难）

方法1：利用group by将s_id，c_id，s_score都重复的去除掉

select a.s_id,a.c_id,a.s_score 
from score a 
join score b on a.s_score=b.s_score 
where a.c_id!=b.c_id and a.s_id=b.s_id 
group by a.s_id,a.c_id,a.s_score;


方法2： 利用distinct将s_id，c_id，s_score都重复的去除掉

select distinct a.s_id,a.c_id,a.s_score 
from score a 
join score b on a.s_score=b.s_score 
where a.c_id!=b.c_id and a.s_id=b.s_id ;

42、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

mysql> select c_id, count(*) count from score group by c_id order by c_id asc;

43、检索至少选修两门课程的学生学号

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

44、查询选修了全部课程的学生信息

方法1 ：

select a.s_id,a.s_name,a.s_birth,a.s_sex 
from student a 
join score b on a.s_id=b.s_id 
group by a.s_id,a.s_name,a.s_birth,a.s_sex 
having count(b.s_id) =(select count(*) from course);


方法2 ：

select * from student  a 
-- 查询score表中s_id=student表中s_id的行数，即统计每个学生的选修课程数量
where (select count(*) from score where s_id = a.s_id)=(select count(*) from course)

45、查询各学生的年龄

-- 时间差函数：timestampdiff
select TIMESTAMPDIFF(SECOND,"2020-02-27",NOW())  -- 31680773
select TIMESTAMPDIFF(MINUTE,"2020-02-27",NOW())  -- 528010
select TIMESTAMPDIFF(HOUR,"2020-02-27",NOW())  -- 8800
select TIMESTAMPDIFF(DAY,"2020-02-27",NOW())  -- 366
select TIMESTAMPDIFF(WEEK,"2020-02-27",NOW())  -- 52
select TIMESTAMPDIFF(MONTH,"2020-02-27",NOW())  -- 12
select TIMESTAMPDIFF(QUARTER,"2020-02-27",NOW())  -- 4
select TIMESTAMPDIFF(YEAR,"2020-02-27",NOW())  -- 1

48、查询两门以上不及格课程的同学的学号及其平均成绩

select s_id,avg(s_score) avg_score 
from score 
where s_score<60 
group by s_id 
having count(*)>=2;

49、查询本月过生日的学生

mysql> select s_id from student where month(s_birth) = month(now());

50、查询下一个月过生日的学生

mysql> select s_id from student where month(s_birth) = month(now())+1;


题目参考：https://blog.csdn.net/qiqi123i/article/details/108455130
题目参考：https://zhuanlan.zhihu.com/p/38354000

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