Mysql经典例题练习与总结
一、创建数据表并插入数据
-- 1、学生表
-- Student(s_id,s_name,s_birth,s_sex) ：学生编号、姓名、年月、性别
CREATE TABLE
IF NOT EXISTS `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`)
) ENGINE = INNODB DEFAULT CHARSET = utf8;
-- 插入数据
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');
-- 2、课程表
-- Course(c_id,c_name,t_id) ：课程编号、 课程名称、 教师编号
CREATE TABLE
IF NOT EXISTS `Course` (
`c_id` VARCHAR (20),
`c_name` VARCHAR (20) NOT NULL DEFAULT '',
`t_id` VARCHAR (20) NOT NULL,
PRIMARY KEY (`c_id`)
) ENGINE = INNODB DEFAULT CHARSET = utf8;
-- 插入数据
INSERT INTO Course VALUES ('01', '语文', '02');
INSERT INTO Course VALUES ('02', '数学', '01');
INSERT INTO Course VALUES ('03', '英语', '03');
-- 3、教师表
-- Teacher(t_id,t_name) ：教师编号、教师姓名
CREATE TABLE
IF NOT EXISTS `Teacher` (
`t_id` VARCHAR (20),
`t_name` VARCHAR (20) NOT NULL DEFAULT '',
PRIMARY KEY (`t_id`)
) ENGINE = INNODB DEFAULT CHARSET = utf8;
-- 插入数据
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
-- Score(s_id,c_id,s_score) ：学生编号、课程编号、分数
CREATE TABLE
IF NOT EXISTS `Score` (
`s_id` VARCHAR (20),
`c_id` VARCHAR (20),
`s_score` INT (3),
PRIMARY KEY (`s_id`, `c_id`)
) ENGINE = INNODB DEFAULT CHARSET = utf8;
-- 插入数据
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);
二、开始解题
2.1 题目1：查询"01"课程比"02"课程成绩高的学生的信息、课程分数
思路: 将01课程成绩和02课程成绩要对比,所以必须要2个成绩表对比,
解法1:三表联合 筛选
解法2:根据学号查询两表对比后符合条件的学生
select * FROM student a,score b,score c
WHERE a.s_id = b.s_id
and a.s_id = c.s_id
and b.c_id = '01'
and c.c_id = '02'
and b.s_score > c.s_score
2.2 题目2：查询平均成绩大于等于60分且总分大于200分的同学且必须考3门的学生编号和学生姓名和平均成绩
思路:
1.必须考三门 成绩表count(s_score) = 3
2.平均成绩大于等于60 avg(s_score) >= 60
注意,这里因为平均成绩可能是循环小数,所以用到保留函数
round(x,y) X是所要修改的值,y表示修改后的小数位数
3.总分大于200 sum(s_score) > 200
三个都是和成绩有关,而且学生编号是主体,显然按学生分组
group by s_id
select
a.s_id,
a.s_name,
ROUND(AVG(b.s_score),2) avg_score,
ROUND(SUM(b.s_score),2) sum_score
from
student a
JOIN score b on a.s_id = b.s_id
GROUP BY
a.s_id
HAVING
avg_score >= 60
and sum_score > 200
and count(b.s_score) = 3;
2.3 题目3：查询平均成绩小于60分的同学的学生编号、学生姓名、平均成绩（包括有成绩的和无成绩）
思路:
1.平均成绩小于60 avg(s_score) < 60
2.包括无成绩 ,ifnull 如果是NULL 则为0 而且无成绩,是每个学生都要有成绩,所以需要学生表左连接右表 left join on
select
a.s_id,
a.s_name,
ROUND(AVG(IFNULL(b.s_score,0)),2) avg_score
from student a
left join score b on a.s_id = b.s_id
group by a.s_id
having
avg_score < 60
题目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
思路:
1.查询学生姓名 student
2.选课总数 count(score.c_id)
3.所有课程总成绩 sum(score.)
select
a.s_id 学生编号,
a.s_name 学生姓名,
COUNT(b.c_id) 选课总数,
SUM(IFNULL(b.s_score,0)) 总成绩
from student a
left join score b on a.s_id = b.s_id
GROUP BY a.s_id
ORDER BY 总成绩 desc
题目5：查询“李”姓老师的数量
思路:like ''李% ’
_代表一个占位符 %代表任意个占位符
select *
from teacher
WHERE teacher.t_name LIKE "李%"
题目6：查询学过张三老师授课的同学信息
思路:
解法1.多表联合查询张三老师授课的学生
解法2. 张三在教师表中,从课程表中找出张三课id,然后和成绩表连接,最后取出学生信息
select a.*
from student a
join score b
on a.s_id = b.s_id
WHERE b.c_id in (
select c.c_id from course c where c.t_id =(
SELECT teacher.t_id from teacher WHERE t_name ='张三'))
select a.*
from
student a
join score b on a.s_id = b.s_id
join course c on b.c_id = c.c_id
join teacher d on c.t_id = d.t_id
WHERE d.t_name = '张三';
题目7：找出没有学过张三老师课程的学生
思路 先将学过张三老师课程的学生找出来,再not in
SELECT t.*
from student t
WHERE t.s_id not in(
select a.s_id
from student a
join score b
on a.s_id = b.s_id
WHERE b.c_id in (
select c.c_id from course c where c.t_id =(
SELECT teacher.t_id from teacher WHERE t_name ='张三'))
)
题目8：查询学过编号为01，并且学过编号为02课程的学生信息
思路:
方法1 : 成绩表自连接, 左表学01课程,右表学02课程,然后学号相等,就是两个都学过的学生 ,然后student查询
方法2 学过02课程的学生 学号 in 学过01的学生学号 ,然后student 查询
select c.*
from score a
join score b on a.s_id = b.s_id and a.c_id = '01' and b.c_id = '02'
join student c on a.s_id = c.s_id;
题目9：查询学过01课程，但是没有学过02课程的学生信息
思路: 学过01课程的学生 not in 学过02课程的学生 ,然后student查询
select a.*
from student a
join score b on a.s_id = b.s_id and b.c_id = '01'
WHERE a.s_id not in (select c.s_id from score c WHERE c.c_id = '02');
题目10：查询没有学完全部课程的同学的信息(本篇内容不考虑重修)
思路: 成绩表中,按学号分组,根据每个学生学习过的课程数量与课程总数对比
select a.*,count(b.c_id) cnt
from student a
join score b on a.s_id = b.s_id
GROUP BY a.s_id
having cnt <(select count(c_id) from course)
题目11：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:
同学学习过的课程只需要有一个在学号01学生学过的课程中即可
c_id in( 01学过的课程)
select a.*
from student a
join score b on a.s_id = b.s_id and a.s_id != '01'
WHERE b.c_id in (select c.c_id from score c where c.c_id = '01')
题目12：查询和01同学学习的课程完全相同的同学的信息
思路:
方法1:首先 个数要相同 其次 同学没学过的课程01没学过
方法2:使用group_concat将列连接,这样就可以直接进行相等了
select a.*,GROUP_CONCAT(b.c_id) course_t
from student a
join score b on b.s_id = a.s_id and a.s_id != '01'
GROUP BY a.s_id
HAVING course_t = (select GROUP_CONCAT(c.c_id) from score c WHERE c.s_id = '01');
题目13：查询没有修过张三老师讲授的任何一门课程的学生姓名
思路: 这个题的话,找出张三老师教授的课程,然后找出学过的学生,not in 学号
select s.*
from student s
WHERE s.s_id not in (
select a.s_id -- 1 2 3 4 5 7
from score a
join course b on a.c_id = b.c_id
join teacher c on c.t_id = b.t_id and c.t_name = '张三')
ORDER BY s_id
题目14：检索01课程分数小于60，按分数降序排列的学生信息
select a.* ,b.s_score
from
student a
join
score b on a.s_id = b.s_id
and b.s_score < 60
and b.c_id = '01'
order by b.s_score desc ;
题目15：按平均成绩从高到低（降序）显示所有学生的所有课程的成绩以及平均成绩
思路:
注意这里要显示所有课程的成绩,以及平均成绩
方法1: 每个课程的成绩都得查询,并且一个课程一个表,就需要原成绩表 3个单列成绩表 4个表,这样太复杂了
方法2: case when
select
a.s_id
,max(case when a.c_id = '01' then a.s_score else 0 end ) 语文
,max(case when a.c_id = '02' then a.s_score else 0 end ) 数学
,max(case when a.c_id = '03' then a.s_score else 0 end ) 英语
,ROUND(AVG(a.s_score)) avgr
from score a
GROUP BY a.s_id
order BY avgr desc
题目16：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（及格：>=60），中等率（中等为：70-80），优良率（优良为：80-90），优秀率（优秀为：>=90）
思路:
根据c_id 分组 然后max(s_score) min(s_score) avg(s_score) case
注意 因为执行顺序 GROUP BY 先于select 所以 select 的每行的数据都是对应group by的,所以这时候对应的每个课程的全部分数
select
a.c_id
,max(a.s_score)
,min(a.s_score)
,round(sum(case when a.s_score >= 60 then 1 else 0 end)/sum(case when a.s_score then 1 else 0 end),2) 及格率
,round(sum(case when a.s_score >=70 and a.s_score <=80 then 1 else 0 end)/sum(case when a.s_score then 1 else 0 end),2) 中等率
,round(sum(case when a.s_score >=80 and a.s_score <=90 then 1 else 0 end)/sum(case when a.s_score then 1 else 0 end),2) 优良率
,round(sum(case when a.s_score >=90 then 1 else 0 end)/sum(case when a.s_score then 1 else 0 end),2) 优良率
from score a
GROUP BY a.c_id
题目17：按照各科成绩进行排序，并且显示排名—比较综合，多看！
思路: 按照各科成绩成绩进行排名, 这个题的话,需要进行union拼接
set @rank = 0;
select
a.c_id
,a.s_score
,(
select COUNT(DISTINCT t.s_score)
from score t
WHERE t.s_score >= a.s_score and a.c_id = t.c_id) 语文排名
from score a
order by a.c_id,a.s_score desc;
题目18：查询学生的总成绩，并进行排名—比较综合，多看！
思路: 先把排名查出来看看,因为要考虑并列情况
按课程号和成绩进行分组,这样就可以按课程和成绩进行组合排名
select
a.c_id
,a.s_score
,(
select COUNT(DISTINCT t.s_score)
from score t
WHERE t.s_score >= a.s_score and a.c_id = t.c_id) 单科排名
from score a
order by a.c_id,a.s_score desc;
题目19：查询学生的总成绩，并进行排名—比较综合，多看！
– 整2个总分表 总分表是一个学生对好几门课的成绩，所以按成绩分组了
– 然后总分表进行比较查出该学生的排名
select t1.s_id,t1.sc1_sum,count(*) AS no
from
(select sc1.s_id,SUM(sc1.s_score) sc1_sum
from score sc1
GROUP BY sc1.s_id) t1
join
(select sc2.s_id,SUM(sc2.s_score) sc2_sum
from score sc2
GROUP BY sc2.s_id) t2 on t1.sc1_sum <= t2.sc2_sum
GROUP BY t1.s_id
order by no;
题目20：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
思路: 先按成绩进行排名, 然后按照limit查询出需要的学生信息
select a1.*,t1.s_score,@cid := '01' c_id
from student a1
join (
select c1.s_id,c1.s_score
from score c1
WHERE c1.c_id = '01'
ORDER BY c1.s_score
limit 1,2
) t1 on a1.s_id = t1.s_id
UNION
select a2.*,t2.s_score,@cid := '02' c_id
from student a2
join (
select c2.s_id,c2.s_score
from score c2
WHERE c2.c_id = '02'
ORDER BY c2.s_score
limit 1,2
) t2 on a2.s_id = t2.s_id
UNION
select a3.*,t3.s_score,@cid := '03' c_id
from student a3
join (
select c3.s_id,c3.s_score
from score c3
WHERE c3.c_id = '03'
ORDER BY c3.s_score
limit 1,2
) t3 on a3.s_id = t3.s_id;
题目21：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
思路: 按课程编号进行分组,然后使用case when 语句分析出每科的相应分段人数
解法只写了一个段的类型占比
select
b.c_id
,b.c_name
,ROUND(SUM(case when a.s_score >= 85 and a.s_score <= 100 then 1 else 0 end)/count(*),2) '100-80'
,ROUND(SUM(case when a.s_score <= 85 and a.s_score >=70 then 1 else 0 end)/count(*),2) '85-70'
,ROUND(SUM(case when a.s_score <= 70 and a.s_score >=60 then 1 else 0 end)/count(*),2) '70-60'
,ROUND(SUM(case when a.s_score <= 60 and a.s_score >=0 then 1 else 0 end)/count(*),2) '0-60'
,count(*)
from score a
join course b on a.c_id = b.c_id
GROUP BY a.c_id;
题目22：查询学生的平均成绩及名次—比较综合，多看，定义变量，实现rank函数
-- 分析 先查询平均成绩
select
a.s_id -- 学号
,@i:=@i+1 as '不保留空缺排名' -- 直接i的自加，顺序一直变大，声明变量需加@ 由于要放入列中需要 :=
,@k:=(case when @i=1 or @avg_score=a.avg_s then @k else @k+1 end) as '保留空缺排名'
-- 因为第一次必定是1所以和i一致为1 后面则按分数是否和上一个相同排名
,@avg_score:=avg_s as '平均分' -- 表a中的值
from (select
s_id
,round(avg(s_score), 2) as avg_s
from Score
group by s_id
order by 2 desc)a -- 表a：平均成绩的排序和学号
,(select @avg_score:=0, @i:=0, @k:=1) b -- 表b：进行变量初始化，固定写法。
-- order by时,把要定义的变量定义在放在后面
本题选自https://blog.csdn.net/qq_40216188/article/details/118670474