我的表是:
id sub_id datetime resource
---|-----|------------|-------
1 | 10 | 04/03/2009 | 399
2 | 11 | 04/03/2009 | 244
3 | 10 | 04/03/2009 | 555
4 | 10 | 03/03/2009 | 300
5 | 11 | 03/03/2009 | 200
6 | 11 | 03/03/2009 | 500
7 | 11 | 24/12/2008 | 600
8 | 13 | 01/01/2009 | 750
9 | 10 | 01/01/2009 | 760
10 | 13 | 01/01/2009 | 570
11 | 11 | 01/01/2009 | 870
12 | 13 | 01/01/2009 | 670
13 | 13 | 01/01/2009 | 703
14 | 13 | 01/01/2009 | 705
我需要为每个sub_id选择2次
结果将是:
id sub_id datetime resource
---|-----|------------|-------
1 | 10 | 04/03/2009 | 399
3 | 10 | 04/03/2009 | 555
5 | 11 | 03/03/2009 | 200
6 | 11 | 03/03/2009 | 500
8 | 13 | 01/01/2009 | 750
10 | 13 | 01/01/2009 | 570
如何在postgres中实现此结果?
答案 0 :(得分:1)
使用窗口函数row_number()
:
select id, sub_id, datetime, resource
from (
select *, row_number() over (partition by sub_id order by id)
from my_table
) s
where row_number < 3;
答案 1 :(得分:1)
查看订单栏(我使用id来匹配您的样本):
t=# with data as (select *,count(1) over (partition by sub_id order by id) from t)
select id,sub_id,datetime,resource from data where count <3;
id | sub_id | datetime | resource
----+--------+------------+----------
1 | 10 | 2009-03-04 | 399
3 | 10 | 2009-03-04 | 555
2 | 11 | 2009-03-04 | 244
5 | 11 | 2009-03-03 | 200
8 | 13 | 2009-01-01 | 750
10 | 13 | 2009-01-01 | 570
(6 rows)