Question

我有下表：

select * from points
+---------+-------------------+------+------+
| NAME    | TITLE             | Type | 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   |
+---------+-------------------+------+------+

所需的输出

+---------+----------+-------+
| NAME    | TITLE    | SCORE |
+---------+----------+-------+
| A       | Hippo    |   60  | <--[(2xH)=40 + (1xM)=20] =60
| B       | Snail    |  100  | <--[(1xH)=70 + (1xM)=20 + (1xL)=10] =100
| C       | Dog      |  100  | <--This should happen because [(1xH)=80 + (1xM)=20] =100
+---------+----------+-------+

需要进行计算：

类型只能有三个值：{H，M，L};
当所有值都存在时，它们的评分如下：

H = 70 M = 20 L = 10
如果某个名称有多种类型（H，M或L），则分配如下：
H /（H的数量）; M /（M的数量）; L /（L的数量）= 100

- 示例：A具有4 H，因此每个H

70/4 = 17.5

但有些名字有一套完整的设置，没有所有'类型。 - 示例：C具有类型值：'H＆amp; M` only

CASE H＆amp; M

H = 80 M = 20

CASE M＆amp; L

M = 60 L = 40

CASE H＆amp; L

H = 90 L = 10

还

如果只有H是presnet H = 100
如果只有M是presnet M = 100
如果只有L是presnet L = 100

Answer 1

如果我理解正确，这就是你想要的：

SELECT name,
       title,
       CAST(
       (      -- only have H, or only have M, or only have L:
         CASE WHEN  `# of H` = 0  AND  `# of M` = 0  THEN  100 * `# of active L` / `# of L`
              WHEN  `# of H` = 0  AND  `# of L` = 0  THEN  100 * `# of active M` / `# of M`
              WHEN  `# of M` = 0  AND  `# of L` = 0  THEN  100 * `# of active H` / `# of H`
              -- only have H & M, or only have H & L, or only have M & L:
              WHEN  `# of H` = 0  THEN  60 * `# of active M` / `# of M` + 40 * `# of active L` / `# of L`
              WHEN  `# of M` = 0  THEN  0  -- ??????????
              WHEN  `# of L` = 0  THEN  80 * `# of active H` / `# of H` + 20 * `# of active M` / `# of M`
              -- have all three:
              ELSE  70 * `# of active H` / `# of H` + 20 * `# of active M` / `# of M` + 10 * `# of active L` / `# of L`
         END
       ) AS SIGNED ) AS score
  FROM ( SELECT name,
                title,
                SUM(IF(         type = 'H', 1, 0))  AS `# of H`,
                SUM(IF(rank AND type = 'H', 1, 0))  AS `# of active H`,
                SUM(IF(         type = 'M', 1, 0))  AS `# of M`,
                SUM(IF(rank AND type = 'M', 1, 0))  AS `# of active M`,
                SUM(IF(         type = 'L', 1, 0))  AS `# of L`,
                SUM(IF(rank AND type = 'L', 1, 0))  AS `# of active L`
           FROM points
          GROUP
             BY name,
                title
       ) t
 ORDER
    BY name
;

MySQL如何更正此子查询？

1 个答案: