Question

我有这张桌子

select * from points

+---------+------+------+
| NAME    | Type | RANK |
+---------+------+------+
| A       | H    |  90  |
| A       | M    | 100  |
| A       | H    | N/A  |
| A       | H    | N/A  |
| A       | H    | N/A  |
| B       | H    | 100  |
| B       | M    | 100  |
| B       | L    | 100  |
| C       | H    |  85  |
| C       | M    | 100  |
+---------+------+------+

我正在使用此查询

SELECT name,
       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  90 * `# of active H` / `# of H` + 20 * `# of active L` / `# of L`
              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,
                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
       ) t
 ORDER
    BY name
;

我得到此输出

+---------+-------+
| NAME    | SCORE |
+---------+-------+
| A       |   60  | <--[(2xH)=40 + (1xM)=20] =60
| B       |  100  | <--[(1xH)=70 + (1xM)=20 + (1xL)=10] =100
| C       |  100  | <--[(1xH)=80 + (1xM)=20] =100
+---------+-------+

我需要这个所需的输出

+---------+-------+
| NAME    | SCORE |
+---------+-------+
| A       |   36  | <--[70/4=(17.5 per H) therefore (17.5)*(rank of that h: 90%)=15.75 + (M values, which equals 20/1 =20 Therefore: rank of that m:100% * 20 = 100) = 36 rounded
| B       |  100  | <--[(1xH)=70 + (1xM)=20 + (1xL)=10] =100
| C       |   88  | <--[(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）

- 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
现在，对于C，类型'H'和'M'必须等于100.

因此，当只有'H`和'M'出现时，它们的评分如下：

H = 80 M = 20

同样，如果另一只动物只带有两个类型值M＆amp; L他们将被评为如下：

M = 60 L = 40

同样，如果另一只动物只带有两个类型值H＆amp; L他们将被评为如下：

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

这看起来很熟悉。： - ）

您的描述在许多地方都不一致 - 例如，您的＆＃34;期望的输出＆＃34; A使用70和20即使A没有类型L - 但如果你的意思是我的意思，那么主要变化您需要将SUM(IF(rank AND type = 'H', 1, 0)) AS `# of active H`更改为SUM(IF(type = 'H', rank / 100.0, 0)) AS `rank of H`（同样适用于M和L），并更改对`# of active H`的所有引用以引用`rank of H`而是。这样每条记录将按其排名的比例包含在内，而不是一个全有或全无的东西。

在将CAST转换为整数时，您还希望使用ROUND代替CAST - 或score -

如何修复此查询以乘以列值

1 个答案: