如何排除具有相似列的行？

Question

有个table:

| date        | action | issue_id |
| 2020-12-14  | close  | 1        |
| 2020-12-15  | close  | 1        |
| 2020-12-16  | close  | 1        |
| 2020-12-14  | close  | 2        |

我如何才能 select 在一个查询中只对每个 issue_id 的最后一行进行操作 == 关闭？

SELECT action, issue_id
FROM table
WHERE action = 'close'
AND ???

Answer 1

仅对这三列进行聚合就足够了：

select issue_id, max(date) as date
from mytable 
where action = 'close'
group by issue_id

如果您需要显示更多列，请使用 distinct on:

select distinct on (issue_id) t.*
from mytable t
where action = 'close'
order by issue_id, date desc

Answer 2

您可以使用 distinct on :

select distinct on (issue_id) t.*
from table t
where action = 'close'
order by issue_id, date desc;

Answer 3

... columns might be null.

所以一定要使用DESC NULLS LAST以避免意外的结果：

SELECT DISTINCT ON (issue_id) *
FROM   tbl
WHERE  action = 'close'
ORDER  BY issue_id, date DESC NULLS LAST;

并且您可能希望将另一个独特的表达式（如 tbl_id）附加到 ORDER BY 列表，以打破相等日期之间的联系（如果可能的话）。否则你会得到一个可以随每次执行而改变的任意选择。

参见：

Select first row in each GROUP BY group?
Sort by column ASC, but NULL values first?

如何排除具有相似列的行？

How to exclude rows with similar columns?

sql

postgresql

datetime

greatest-n-per-group