MySQL我将如何处理这种类型的点系统？_Mysql

MySQL我将如何处理这种类型的点系统？

mysql

MySQL我将如何处理这种类型的点系统？,mysql,Mysql,从上面提取的精确数据示例 +---------+-------------------+------+------+ | NAME | TITLE | SIZE | RANK | +---------+-------------------+------+------+ | A | Hippo | H | 1 | | A | Hippo | M | 1 | | A

从上面提取的精确数据示例

+---------+-------------------+------+------+
| NAME    | TITLE             | SIZE | RANK |
+---------+-------------------+------+------+
| A       | Hippo             | H    |  1   |
| A       | Hippo             | M    |  1   |
| A       | Hippo             | H    | N/A  |
| A       | Hippo             | H    |  1   |
| A       | Hippo             | H    | N/A  |
| B       | Snail             | H    |  1   |
| B       | Snail             | M    |  1   |
| B       | Snail             | L    |  1   |
| C       | Dog               | H    |  1   |
| C       | Dog               | M    |  1   |
+---------+-------------------+------+------+

要点如下：

+---------+-------------------+------+------+
| A       | Hippo             | H    |  1   |
| A       | Hippo             | M    |  1   |
| A       | Hippo             | H    | N/A  |
| A       | Hippo             | H    |  1   |
| A       | Hippo             | H    | N/A  |
+---------+-------------------+------+------+

河马在尺寸栏中有4个
```
H
```
```
H
```
总计可以等于70
因此，70/4=17.5
这意味着每个
```
H
```
只能得到17.5分
如果存在“等级”，则获得分数
河马现在只有两个“等级”
因此河马获得（17.5）x（2）=35分
Hippo在“尺寸”栏中还有一个
```
M
```
因此
```
M
```
=20/1=20
因此河马获得20分
河马在尺寸栏中没有
```
L
```
因此没有得分
总共35（从
```
H
```
）+20（从
```
M
```
）+0（从
```
L
```
）=55

期望输出

H = 70 points
M = 20 points
L = 10 points

在复杂的子选择和联接之后，如何在MySQl中进行这种复杂的评分/变量处理，下面是您期望的结果：

+---------+----------+-------+
| NAME    | TITLE    | SCORE |
+---------+----------+-------+
| A       | Hippo    |   55  |
| B       | Snail    |  100  |
| C       | Dog      |   90  |
+---------+----------+-------+

select c.name,c.title,sum(score) from (
select b.name, b.title, avg_point*needcount as score from (select name, title, size,    count(size),  (case when size='H' then 70 when size='M' then 20 when size='L' then 10 end )/ count(size)  as avg_point from points group by name,title,size ) a
join 
(select name, title, size, count(size) as needcount from points where rank group by name, title, size) b
on a.name = b.name
and a.title = b.title
and a.size = b.size) c group by c.name,c.title;