Fork me on GitHub

MySQL50-7-第21-25题

Posted on | In , |
Words count in article: 3.3k | Reading time ≈ 16

MySQL50-7-第11-25题

本文中介绍的是第11-25题目，主要涉及的知识点是：

  • 分组统计求和与百分比
  • 如何利用SQL实现排序
  • having使用
  • union拼接

5个题目分别是

  • 查询不同老师所教不同课程平均分从高到低显示
  • 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
  • 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
  • 查询学生的平均成绩及名次
  • 查询各科成绩前三名的记录

题目21

题目需求

查询不同老师所教不同课程平均分从高到低显示

分析过程

涉及到的表主要是

老师：Teacher

课程：Course，作为主表

成绩：Score

通过3个表的连接求出来即可

SQL实现

先找出每个老师教授了哪些课程：

1
2
3
4
5
6
select
	c.c_name
	,t.t_name
from Course c
left join Teacher t
on c.t_id = t.t_id;

将上面的结果和成绩表连接起来：

1
2
3
4
5
6
7
8
9
10
11
select
	c.c_name
	,t.t_name
	,round(avg(s.s_score),2)  score   -- 课程分组后再求均值
from Course c   -- 主表，通过两次连接
left join Teacher t
on c.t_id = t.t_id
left join Score s
on c.c_id = s.c_id
group by c.c_id   -- 课程分组
order by 3 desc;  -- 降序

题目22

题目需求

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

分析过程

成绩：Score

学生信息：Student

我们通过取出每科的第2、3名拼接起来再取出学生信息

SQL实现

自己的方法

1、课程表和成绩表连接起来，显示所有的课程和成绩信息

1
2
3
4
5
6
7
8
select
	s.s_id
	,s.c_id
	,s.s_score
	,c.c_name
from Score s
join Course c
on s.c_id = c.c_id

2、查出全部的语文成绩

1
2
3
4
5
select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '语文'
order by s.s_score desc;

3、我们找出语文的第2、3的学生

1
2
3
4
5
6
select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '语文'
order by s.s_score desc
limit 1, 2;

4、同时求出语文、数学、英语的分数，并且通过union拼接

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
-- union连接

(select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '语文'
order by s.s_score desc
limit 1, 2)

union

(select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '数学'
order by s.s_score desc
limit 1, 2)

union
((select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '英语'
order by s.s_score desc
limit 1, 2))

5、将上面的结果学生信息表进行连接即可

好歹是实现了😭

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
-- 最终脚本
-- ！！！！真的需要好好优化下

select
	s.s_id
	,s.s_name
	,t.c_name
	,t.s_score
from Student s
join (-- union连接

(select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '语文'
order by s.s_score desc
limit 1, 2)

union

(select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '数学'
order by s.s_score desc
limit 1, 2)

union
((select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '英语'
order by s.s_score desc
limit 1, 2)))t  -- 临时表t

on s.s_id = t.s_id

和第25题相同的方法

1、以语文为例，首先我们找出前3名的成绩（包含相同的成绩）

1
2
3
4
5
6
7
8
9
10
11
12
13
14
-- 语文
select
	a.s_id
	,a.c_id
	,a.s_score   -- 3、此时a表的成绩就是我们找的
from Score a
join Score b
on a.c_id = b.c_id
and a.s_score <= b.s_score  -- 1、判断a的分数小于等于b的分数，要带上等号
and a.c_id="01"
group by 1,2
having count(b.s_id) <= 3  -- 2、b中的个数至少有3个，应对分数相同的情形
order by 3 desc
limit 1,2

1
2
3
4
5
6
7
8
9
10
11
12
13
14
-- 语文
select
	a.s_id
	,a.c_id
	,a.s_score   -- 3、此时a表的成绩就是我们找的
from Score a
join Score b
on a.c_id = b.c_id
and a.s_score <= b.s_score  -- 1、判断a的分数小于等于b的分数，要带上等号
and a.c_id="01"
group by 1,2
having count(b.s_id) <= 3  -- 2、b中的个数至少有3个，应对分数相同的情形
order by 3 desc
limit 1,2;   -- 取得第2、3名

在通过数学和英语的类似操作得到2、3名的成绩，再进行拼接即可

题目23

题目需求

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

分析过程

课程：Course

成绩：Score

通过case语句来进行判断，count语句进行统计，sum语句进行求和

SQL实现

自己的方法

1、如何对每个成绩进行分组展示：ABCD代表相应的等级

1
2
3
4
5
6
7
8
9
select
	c_id
	,s_score
	,case when s_score >= 85 and  s_score<= 100 then 'A'   -- 大小关系必须分两次写，一次写的话MySQL无法识别
		when 70 <= s_score and s_score < 85 then 'B'
		when 60 <= s_score and s_score < 70 then 'C'
		when 0 <= s_score and s_score < 60 then 'D'
		else '其他' end as 'category'
from Score s;

2、将两个表关联起来展示数据

1
2
3
4
5
6
7
8
9
10
11
12
13
14
-- 1、查看全部课程和成绩信息

select
	s.c_id
	,c.c_name
	,s.s_score
	,case when s.s_score >= 85 and  s.s_score<= 100 then 'A'   -- 大小关系必须分两次写，一次写的话MySQL无法识别
		when 70 <= s.s_score and s.s_score < 85 then 'B'
		when 60 <= s.s_score and s.s_score < 70 then 'C'
		when 0 <= s.s_score and s.s_score < 60 then 'D'
		else '其他' end as 'category'
from Score s
join Course c
on s.c_id = c.c_id;

3、完整代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
select
	s.c_id  编号
	,c.c_name  科目
	,sum(case when s.s_score >= 85 and  s.s_score<= 100 then 1 else 0 end) "[85,100]人数"
	,round(100 * (sum(case when s.s_score >= 85 and  s.s_score<= 100 then 1 else 0 end) / sum(case when s.s_score then 1 else 0 end)), 2) as '[85,100]百分比'
	,sum(case when s.s_score >= 70 and  s.s_score<= 85 then 1 else 0 end) "[70,85]人数"
	,round(100 * (sum(case when s.s_score >= 70 and  s.s_score<= 85 then 1 else 0 end) / sum(case when s.s_score then 1 else 0 end)), 2) as '[70,85]百分比'
	,sum(case when s.s_score >= 60 and  s.s_score<= 70 then 1 else 0 end) "[60,70]人数"
	,round(100 * (sum(case when s.s_score >= 60 and  s.s_score<= 70 then 1 else 0 end) / sum(case when s.s_score then 1 else 0 end)), 2) as '[60,70]百分比'
	,sum(case when s.s_score >= 0 and  s.s_score<= 60 then 1 else 0 end) "[0,60]人数"
	,round(100 * (sum(case when s.s_score >= 0 and  s.s_score<= 60 then 1 else 0 end) / sum(case when s.s_score then 1 else 0 end)), 2) as '[0,60]百分比'
from Score s
left join Course c
on s.c_id = c.c_id
group by s.c_id, c.c_name

参考方法

1、先统计每个阶段的人数和占比

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
select
	c_id
	,sum(case when s_score > 85 and s_score <=100 then 1 else 0 end) as '85-100'
	,round(100 * (sum(case when s_score > 85 and s_score <= 100 then 1 else 0 end) / count(*)), 2) '占比'
from Score
group by c_id;  -- 分课程统计总数和占比


-- 方式2
select
	c_id
	,sum(case when s_score > 85 and s_score <=100 then 1 else 0 end) as '85-100'
	,round(100 * (sum(case when s_score > 85 and s_score <= 100 then 1 else 0 end) / count(case when s_score then 1 else 0 end)), 2) '占比'   -- 不同count(*)
from Score
group by c_id;

注意对比：

2、我们将4种情况同时查出来

1
2
3
4
5
6
7
8
9
10
11
12
select
	c_id
	,sum(case when s_score > 85 and s_score <=100 then 1 else 0 end) as '85-100'
	,round(100 * (sum(case when s_score > 85 and s_score <= 100 then 1 else 0 end) / count(*)), 2) '[85,100]占比'
	,sum(case when s_score > 70 and s_score <=85 then 1 else 0 end) as '70-85'
	,round(100 * (sum(case when s_score > 70 and s_score <= 85 then 1 else 0 end) / count(*)), 2) '[70,85]占比'
	,sum(case when s_score > 60 and s_score <=70 then 1 else 0 end) as '60-70'
	,round(100 * (sum(case when s_score > 60 and s_score <= 70 then 1 else 0 end) / count(*)), 2) '[60,70]占比'
	,sum(case when s_score >=0 and s_score <=60 then 1 else 0 end) as '0-60'
	,round(100 * (sum(case when s_score > 0 and s_score <= 60 then 1 else 0 end) / count(*)), 2) '[0,60]占比'
from Score
group by c_id;  -- 分课程统计总数和占比

3、将科目名称连接起来

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
-- 整体和自己的方法是类似的
select
	s.c_id
	,c.c_name
	,sum(case when s_score > 85 and s_score <=100 then 1 else 0 end) as '85-100'
	,round(100 * (sum(case when s_score > 85 and s_score <= 100 then 1 else 0 end) / count(*)), 2) '[85,100]占比'
	,sum(case when s_score > 70 and s_score <=85 then 1 else 0 end) as '70-85'
	,round(100 * (sum(case when s_score > 70 and s_score <= 85 then 1 else 0 end) / count(*)), 2) '[70,85]占比'
	,sum(case when s_score > 60 and s_score <=70 then 1 else 0 end) as '60-70'
	,round(100 * (sum(case when s_score > 60 and s_score <= 70 then 1 else 0 end) / count(*)), 2) '[60,70]占比'
	,sum(case when s_score >=0 and s_score <=60 then 1 else 0 end) as '0-60'
	,round(100 * (sum(case when s_score > 0 and s_score <= 60 then 1 else 0 end) / count(*)), 2) '[0,60]占比'
from Score s
left join Course c
on s.c_id = c.c_id
group by s.c_id, c.c_name;    -- 分课程统计总数和占比

题目24

题目需求

查询学生的平均成绩及名次

分析过程

学生：Student

成绩：Score

平均：avg函数

名次：通过排序来解决

SQL实现

自己的方法

1、先求出每个人的平均分

1
2
3
4
5
6
7
8
9
10
-- 自己的方法

select
	sc.s_id
	,s.s_name
	,round(avg(sc.s_score),2)  avg_score
from Score sc
join Student s
on sc.s_id=s.s_id
group by sc.s_id,s.s_name

2、我们对上面的结果进行排序

！！！MySQL5中是没有rank函数的，需要自己实现排序功能

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
-- MYSQL5.7中没有rank函数，所以通过自连接实现

select
	t1.s_id
	,t1.s_name
	,t1.avg_score
	,(select count(distinct t2.avg_score)
		from (select
            sc.s_id
            ,s.s_name
            ,round(avg(sc.s_score),2)  avg_score
          from Score sc
          join Student s
          on sc.s_id=s.s_id
          group by sc.s_id,s.s_name)t2    -- 临时表t2也是上面的结果
		where t2.avg_score >= t1.avg_score
		) rank

from (select
        sc.s_id
        ,s.s_name
        ,round(avg(sc.s_score),2)  avg_score
      from Score sc
      join Student s
      on sc.s_id=s.s_id
      group by sc.s_id,s.s_name)t1   -- 临时表t1就是上面的结果
order by t1.avg_score desc;

参考方法
1
2
3
4
5
6
7
8
9
10
11
12
13
select
	a.s_id  -- 学号
	,@i:=@i+1 as '不保留空缺排名'   -- 直接i的自加
	,@k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名'
	,@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:=0)b   -- 表b：通过变量设置初始值

实现rank函数

1
2
3
4
5
6
7
8
select
	s.s_name  -- 姓名
	,s.s_score  -- 成绩
	,(select count(distinct t2.s_score)
    from Score t2
    where t2.s_score >= t1.s_score) rank   -- 在t2分数大的情况下，统计t2的去重个数
from Score t1
order by t1.s_score desc;   -- 分数降序排列

举例子来说明这个脚本：

姓名 成绩
张三 89
李四 90
王五 78
小明 98
小红 60
  1. 当t1.s_score=89，满足t2.s_score > = t1.s_score的有98，90和89，此时count(distinct t2.s_score) 的个数就是3
  2. 当t1.s_score=90，满足t2.s_score > = t1.s_score的有98和90，此时count(distinct t2.s_score) 的个数就是2
  3. 当t1.s_score=78，满足t2.s_score > = t1.s_score的有98、90、89和78，此时count(distinct t2.s_score) 的个数就是4
  4. 当t1.s_score=98，满足t2.s_score > = t1.s_score的只有98，此时count(distinct t2.s_score) 的个数就是1
  5. 当t1.s_score=60，满足t2.s_score > = t1.s_score的有89、90、78、98、60，此时count(distinct t2.s_score) 的个数就是5

通过上面的步骤，我们发现：t1中每个分数对应的个数就是它的排名

题目25

题目需求

查询各科成绩前三名的记录

分析过程

这题和第22题是属于一个类型的：找到每个科目的指定名次的成绩，使用的表是：Score

SQL实现

自己的方法

1、首先我们找出语文的前3名

1
2
3
4
5
6
select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '语文'
order by s.s_score desc   -- 降序之后取出前3条记录
limit 3;

2、通过同样的方法我们可以求出数学和英语的前3条记录，然后通过union进行联结，有待优化😭

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
-- 自己的脚本

(select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '语文'
order by s.s_score desc   -- 降序之后取出前3条记录
limit 3)

union

(select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '数学'
order by s.s_score desc
limit 3)

union

(select s.s_id, s.s_score, c.c_name
from Score s
join Course c on s.c_id = c.c_id
where c.c_name = '英语'
order by s.s_score desc
limit 3)

参考方法

通过Score表的自连接，表a中的值小于表b中的值，排序之后我们取前3

1
2
3
4
5
6
7
8
9
10
11
select
	a.s_id
	,a.c_id
	,a.s_score   -- 取出a中的成绩
from Score a
join Score b
on a.c_id = b.c_id
and a.s_score <= b.s_score   -- 表b中的成绩大
group by 1,2,3
having count(b.s_id) = 3
order by 2, 3 desc;

我们通过语文这个科目来理解上面的代码：前3名是80，80，76

1
2
3
4
5
6
7
8
9
10
11
12
13
-- 语文
select
	a.s_id
	,a.c_id
	,a.s_score   -- 3、此时a表的成绩就是我们找的
from Score a
join Score b
on a.c_id = b.c_id
and a.s_score <= b.s_score  -- 1、判断a的分数小于等于b的分数，要带上等号
and a.c_id="01"
group by 1,2
having count(b.s_id) <= 3  -- 2、b中的个数至少有3个，应对分数相同的情形
order by 3 desc;

1
2
3
4
5
6
7
8
9
10
11
12
13
-- 语文

select
	a.s_id
	,a.c_id
	,a.s_score   -- a表的成绩
from Score a
join Score b
on a.c_id = b.c_id
and a.s_score <= b.s_score   -- 1、判断a的分数小于等于b的分数，要带上等号
group by 1,2,3
having count(b.s_id) <= 3   -- 2、b中的个数至少有3个，应对分数相同的情形
order by 2, 3 desc;   -- 课程（2）的升序，成绩（）3的降序

本文标题:MySQL50-7-第21-25题

发布时间:2020年11月21日 - 17:11

原始链接:http://www.renpeter.cn/2020/11/21/MySQL50-7-%E7%AC%AC21-25%E9%A2%98.html

许可协议: 署名-非商业性使用-禁止演绎 4.0 国际 转载请保留原文链接及作者。

Coffee or Tea