文章目录 1. 数据准备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）2、查询平均成绩大于60分的学生的学号和平均成绩3、查询所有学生的学
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)
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);数据库表：
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
(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
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;
mysql> select count(t_name) from teacher where t_name='猴';
select s_id,s_name from student where s_id not in (select a.s_idfrom score a join student b on a.s_id=b.s_idjoin course c on a.c_id=c.c_idjoin teacher d on c.t_id=d.t_idwhere 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_idfrom score a join student b on a.s_id=b.s_idjoin course c on a.c_id=c.c_idjoin teacher d on c.t_id=d.t_idwhere 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
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';
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 ajoin (select s_id from score where c_id='01') b on a.s_id=b.s_idjoin (select s_id from score where c_id='02') c on a.s_id=c.s_idjoin 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
mysql> select sum(s_score) from score group by c_id having c_id='02';
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
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_namehaving count(b.c_id)<(select count(c_id) from course);
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
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');
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');
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
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;
select a.s_id,a.s_score,b.avg_scorefrom 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_scorefrom 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
(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 join 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 ajoin course bon a.c_id=b.c_idgroup by a.c_id;
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;
select c_id,avg(s_score) avg_score from score group by c_id order by avg_score desc;
select s_id,avg(s_score) avg_score,row_number () over( order by avg(s_score) desc) row_numfrom score group by s_id;
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;
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;
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;
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;
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;
mysql> select c_id,count(c_id) count from score group by c_id;
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;
mysql> select count(s_sex) count_sex from student group by s_sex;
mysql> select * from student where s_name like '%周%';
mysql> select s_id,s_name from student where year(s_birth)=1990;
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;
select c_id, avg(s_score) avg_score from score group by c_id order by avg_score,c_id desc;
select a.c_id,b.c_name,c.s_id,c.s_name,a.s_scorefrom 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';
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 ;
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;
mysql> select distinct c_id from score where s_score<60 order by c_id desc;
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;
mysql> select c_id, count(*) count from score group by c_id;
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;
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 ;
mysql> select c_id, count(*) count from score group by c_id order by c_id asc;
select s_id,count(*) count from score group by s_id having count>=2;
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) 
-- 时间差函数：timestampdiffselect TIMESTAMPDIFF(SECOND,"2020-02-27",NOW()) -- 31680773select TIMESTAMPDIFF(MINUTE,"2020-02-27",NOW()) -- 528010select TIMESTAMPDIFF(HOUR,"2020-02-27",NOW()) -- 8800select TIMESTAMPDIFF(DAY,"2020-02-27",NOW()) -- 366select TIMESTAMPDIFF(WEEK,"2020-02-27",NOW()) -- 52select TIMESTAMPDIFF(MONTH,"2020-02-27",NOW()) -- 12select TIMESTAMPDIFF(QUARTER,"2020-02-27",NOW()) -- 4select TIMESTAMPDIFF(YEAR,"2020-02-27",NOW()) -- 1
select s_id,avg(s_score) avg_score from score where s_score<60 group by s_id having count(*)>=2;
mysql> select s_id from student where month(s_birth) = month(now());
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
来源地址：https://blog.csdn.net/qq_42764468/article/details/122294619
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本文标题: 深耕MySQL - 50道SQL练习题
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