如何按多行多列分组

Question

1. 每个用户有 3 个代码 (10,20,30)
2. 如果所有代码（10,20,30）都为真，则必须加入
3. 如果任何代码 (10,20,30) 为 False，则必须忽略
4. 我们必须检查最大值（日期）

表格如下：

+-------+------+-------+------------+
| user  | code | bool  |    date    |
+-------+------+-------+------------+
| user1 |  10  | True  | 31/10/2019 |
| user1 |  20  | True  | 31/10/2019 |
| user1 |  30  | False | 31/10/2019 |
| user2 |  10  | True  | 31/10/2019 |
| user2 |  20  | True  | 31/10/2019 |
| user2 |  30  | True  | 31/10/2019 |
| user3 |  10  | True  | 31/10/2019 |
| user3 |  20  | True  | 31/10/2019 |
| user3 |  30  | True  | 31/10/2019 |
+-------+------+-------+------------+

输出 user2 和 user3 已加入，user1 已被忽略

+------------+--------+-----------+
|    DATE    | JOINED | NEGLECTED |
+------------+--------+-----------+
| 31/10/2019 |    2   |    1      |
+------------+--------+-----------+

Answer 1

我假设列bool 的数据类型为BOOL，因此您可以在2 个聚合级别中进行：

select t.date,
  sum(total = 3) joined,
  sum(total <> 3) neglected
from (
  select date, user, sum(bool) total
  from tablename
  group by date, user
) t
group by t.date

请参阅demo。
结果：

| date       | joined | neglected |
| ---------- | ------ | --------- |
| 31/10/2019 | 2      | 1         |

如果列的bool 数据类型为VARCHAR，则更改：

sum(bool) total

到：

sum(bool = 'True') total

Answer 2

我将从一个查询开始，该查询将用户分组并根据代码计算他们的状态。然后，我会用一个聚合查询来包装这个查询，以计算每天加入或忽略的用户数：

SELECT   `date`,
         SUM(`status` = 'joined') AS joined,
         SUM(`status` = 'neglected') AS neglected
FROM     (SELECT   `user`,
                   `date`,
                   CASE BIT_AND(bool) WHEN TRUE THEN 'joined' ELSE 'neglected' END AS status
          FROM     `mytable`
          GROUP BY `user`, `date`) t
GROUP BY `date`