获取具有条件的行

Question

每个客户可以拥有一个或多个帐户（account_id）

如何为关闭所有帐户的客户和其余活跃客户获得最大值 closed_date？*

*活跃客户是指至少拥有一个开立账户的客户。

+-------------+------------+--------------+-------------+
| customer_id | account_id | created_date | closed_date |
+-------------+------------+--------------+-------------+
| 3eba3       | 5dddd      | 17/06/2020   |             |
| 3eba3       | eabbd      | 29/06/2020   |             |
| 3eba3       | 9f3a4      | 29/06/2020   | 09/11/2020  |
| 5hlf1       | khti1      | 01/02/2020   |             |
| hdk12       | sfsf2      | 05/03/2020   | 01/06/2020  |
| hdk12       | sfsl3      | 06/03/2020   | 01/10/2020  |
| 12kju       | gege1      | 07/03/2020   | 01/07/2020  |
| 12kju       | mhfl1      | 08/03/2020   | 03/07/2020  |
+-------------+------------+--------------+-------------+

期望输出：

+-------------+-------------+
| customer_id | closed_date |
+-------------+-------------+
| 3eba3       |             |
| 5hlf1       |             |
| hdk12       | 01/10/2020  |
| 12kju       | 03/07/2020  |
+-------------+-------------+ 

Answer 1

您可以为此使用 distinct on：

select distinct on (customer_id) t.*
from t
order by customer_id, closed_date desc nulls first;

这 returns 每 customer_id 一行基于 order by 子句。

Answer 2

您可以使用聚合和 case 表达式：

select customer_id,
    case when bool_and(closed_date is not null) then max(closed_date) end as closed_date
from mytable
group by customer_id