Question

我有这样的观点，这是由于多重联接造成的：

project_id  | document_type_id
10          | 2
10          | 2
10          | 3
10          | 1
10          | 1
10          | 1
11          | 2
11          | 2
11          | 2
11          | 2
11          | 3
11          | 3

标签1：“评论”

标签2：“面试”

3的标签：“浪漫”

我已经获得了这张桌子：

project_id  | review     | interview | romance
10          | OK         | OK        | OK
11          | NO         | OK        | OK

使用以下查询语句：

SELECT `project_id`, `document_type_id`,
MAX(CASE WHEN `document_type_id` = 1 THEN "OK" ELSE "NO" END) as "review",
MAX(CASE WHEN `document_type_id` = 2 THEN "OK" ELSE "NO" END) as "interview",
MAX(CASE WHEN `document_type_id` = 3 THEN "OK" ELSE "NO" END) as "romance"

FROM projectDocumentList
GROUP BY project_id

我现在需要的是计算像这样的表中每个字段中的每种文档类型：

project_id  | review | interview | romance
10          | 3      | 2         | 1
11          | 0      | 4         | 2

我尝试了很多次，但是我找不到正确的正税来获得此结果。我以前的一些尝试...

COUNT(MAX(CASE WHEN `document_type_id` = 1 THEN "1" ELSE "NO" END)) as "review"

MAX(CASE WHEN `document_type_id` = 1 THEN (SELECT COUNT(`document_type_id`)) ELSE "NO" END) as "review"

COUNT(DISTINCT(MAX(CASE WHEN `document_type_id` = 1 THEN `document_type_id` ELSE "NO" END)) as "review"

有关任何想法？谢谢。 V。

Answer 1

您可以使用SUM()

查询

SELECT `project_id`,
SUM(CASE WHEN `document_type_id` = 1 THEN 1 ELSE 0 END) as "review",
SUM(CASE WHEN `document_type_id` = 2 THEN 1 ELSE 0 END) as "interview",
SUM(CASE WHEN `document_type_id` = 3 THEN 1 ELSE 0 END) as "romance"
FROM projectDocumentList
GROUP BY project_id

结果

| project_id | review | interview | romance |
| ---------- | ------ | --------- | ------- |
| 10         | 3      | 2         | 1       |
| 11         | 0      | 4         | 2       |

demo

也可以使用COUNT(..)，但是您需要使用NULL而不是0。
这是因为COUNT(..)处理0, 1 and NULL values的方式与SUM(..)不同。

查询

SELECT `project_id`,
COUNT(CASE WHEN `document_type_id` = 1 THEN 1 ELSE NULL END) as "review",
COUNT(CASE WHEN `document_type_id` = 2 THEN 1 ELSE NULL END) as "interview",
COUNT(CASE WHEN `document_type_id` = 3 THEN 1 ELSE NULL END) as "romance"
FROM projectDocumentList
GROUP BY project_id

结果

| project_id | review | interview | romance |
| ---------- | ------ | --------- | ------- |
| 10         | 3      | 2         | 1       |
| 11         | 0      | 4         | 2       |

demo

COUNT MAX CASE和GROUP BY用于重复的类型ID

1 个答案: