MySQL如何实现这个查询？

Question

我有以下模拟表：

+---------+-------------------+------+------+
| NAME    | TITLE             | SIZE |  Hit |
+---------+-------------------+------+------+
| A       | Hippo1            | H    |  0   |
| A       | Hippo2            | H    |  0   |
| A       | Hippo3            | H    |  1   |   
| A       | Hippo1            | M    |  0   |
| A       | Hippo2            | M    |  1   |
| A       | Hippo3            | M    |  1   |
| A       | Hippo1            | L    |  0   |
| A       | Hippo2            | L    |  1   |    
| A       | Hippo3            | L    |  1   |
+---------+-------------------+------+------+
| B       | Snail1            | H    |  1   |
| B       | Snail1            | M    |  0   |
| B       | Snail1            | L    |  1   |
+---------+-------------------+------+------+
| C       | Dog               | H    |  1   |
| C       | Dog               | M    |  0   |
+---------+-------------------+------+------+
| D       | Sheep             | H    |  0   |
| D       | Sheep             | L    |  1   |
+---------+-------------------+------+------+
| E       | Fish              | H    |  1   |
| E       | Fish              | H    |  1   |
+---------+-------------------+------+------+

我想要的结果

+---------+----------+-------+
| NAME    | TITLE    | SCORE |
+---------+----------+-------+
| A       | Hippo    |  100  |
| B       | Snail    |   80  |
| C       | Dog      |   80  |
| D       | Sheep    |   10  |
| E       | Fish     |  100  |
+---------+----------+-------+

需要计算：

• 类型最多只能有三个值：{H, M, L};
• 当所有值都存在时，它们的等级如下：H=70 M=20 L=10

• 所有独特的案例都是

  1. 案例 {H,M}：H=80 M=20
  2. 案例 {M,L}：M=60 L=40
  3. 案例 {H,L}：H=90 L=10
  4. 案例 {H}：H=100
  5. 案例 {M}：M=100
  6. 案例 {L}：L=100
  7. 案例{H,M,L}：H=70 M=20 L=10

说明：

Hippo Has case {H,M,L}
Snail has case {H,M,L}
Dog has case {H,M}
Sheep has case {H,L}
Fish has case {H} 

进一步说明

Hippo ：尽管并非所有尺寸都有成功，但 Hippo 获得了 100 分，因为每个案例中至少有 1 个已满足每个标题一次。因此 HIPPO3 拥有所有三个 H M Lsatsified 所以 hippo 可以认为是 100% 找到

从上面提取（行是查看器的逻辑构造）

| A       | Hippo1            | H    |  0   |
| A       | Hippo2            | H    |  0   |
| A       | Hippo3            | H    |  1   | <--Here   
+-------------------------------------------+    
| A       | Hippo1            | M    |  0   |
| A       | Hippo2            | M    |  1   |
| A       | Hippo3            | M    |  1   | <--Here
+-------------------------------------------+   
| A       | Hippo1            | L    |  0   |
| A       | Hippo2            | L    |  1   |    
| A       | Hippo3            | L    |  1   | <--Here

Answer 1

给你：

select name, sum(FinalVal) Score from (
  select distinct t.name,
    case
      when size = 'H' then Hval
      when size = 'M' then Mval
      else Lval
    end FinalVal
  from (
    select name,
      case
        when sizes = 'H,L,M' then 70
        when sizes = 'H,M' then 80
        when sizes = 'H,L' then 90
        when sizes = 'H' then 100
        else 0
      end Hval,
      case
        when sizes = 'H,L,M' then 20
        when sizes = 'H,M' then 20
        when sizes = 'L,M' then 60
        when sizes = 'M' then 100
        else 0
      end Mval,
      case
        when sizes = 'H,L,M' then 10
        when sizes = 'L,M' then 40
        when sizes = 'H,L' then 10
        when sizes = 'L' then 100
        else 0
      end Lval
    from (
      select name, group_concat(distinct size order by size) sizes from t
      group by name
    ) s1
  ) s2
  join t on t.name = s2.name
  where hit
) final
group by name

不清楚如何根据您的数据获取标题（实际上结果具有原始表中不存在的数据，例如Hippo，而不是Hippo1。但我会把它留给你.

这将导致：

+------+-------+
| NAME | SCORE |
+------+-------+
| A    |   100 |
| B    |    80 |
| C    |    80 |
| D    |    10 |
| E    |   100 |
+------+-------+