SQL将枚举创建为列并按日期分组

Question

我有一张像下面这样的表格

id	folder_num	category	date
1	100/2021	add	8/25/2021 20:58
2	200/2021	delete	8/25/2021 21:02
3	300/2021	add	8/25/2021 21:06
4	400/2021	add	8/25/2021 21:10
5	500/2021	delete	8/26/2021 21:14
6	600/2021	edit	8/26/2021 21:18
7	700/2021	edit	8/27/2021 21:22
8	800/2021	edit	8/27/2021 21:26
9	900/2021	edit	8/27/2021 21:30
10	1000/2021	add	8/27/2021 21:34

列category是一个枚举类型的列，我想把枚举的内容做成一个列并按日期分组。

我之前根据category 和date 创建了一个分组，结果如下。

SELECT c.category, 
        COUNT(l.id) total,  
        d.date as date
 FROM (SELECT DISTINCT category FROM log) c
 CROSS JOIN (SELECT DISTINCT DATE(date) date FROM log WHERE DATE(date) BETWEEN '2021-08-01' AND '2021-08-30') d
 LEFT JOIN log l ON l.category = c.category AND DATE(l.date) = d.date
 GROUP BY c.category, d.date
 ORDER BY c.category ASC

category	total	date
add	3	8/25/2021
delete	1	8/25/2021
edit	0	8/25/2021
add	0	8/26/2021
delete	1	8/26/2021
edit	1	8/26/2021
add	1	8/27/2021
delete	0	8/27/2021
edit	3	8/27/2021

结果还算满意，但回到我的问题，我想把category列中的值变成一列而不是一行，结果是这样的

date	add	delete	edit
8/25/2021	3	1	0
8/26/2021	0	1	1
8/27/2021	1	0	3

我尝试使用 max(case ...) 函数但代码错误，这是我尝试过的

SELECT 
        d.date as date,
        max(case when d.category = 'add' then COUNT(l.id) else null end) as add,
        max(case when d.category = 'delete' then COUNT(l.id) else null end) as delete,
        max(case when d.category = 'edit' then COUNT(l.id) else null end) as edit
 FROM (SELECT DISTINCT category FROM log) c
 CROSS JOIN (SELECT DISTINCT DATE(date) date FROM log WHERE DATE(date) BETWEEN '2021-08-01' AND '2021-08-30') d
 LEFT JOIN log l ON l.category = c.category AND DATE(l.date) = d.date
 GROUP BY d.date

我的查询中函数的使用是否放错了位置？谢谢

Answer 1

你不能嵌套聚合函数，但你可以简单地使用：

SELECT date(l.date) as date,
       sum(l.category = 'add') as add,
       sum(l.category = 'delete') as delete,
       sum(l.category = 'edit') as edit
FROM log l
WHERE l.date >= '2021-08-01' AND l.date < '2021-08-31'
GROUP BY l.date