在postgres查询中获取第一行 [英] getting first row in postgres query

问题描述

我正在使用内部联接从2个表中查询一些数据.

I am querying some data from 2 tables using inner join.

这是查询

test_db=> select api_booking.install_ts, api_user.id from api_booking inner join api_user on api_booking.user_id=api_user.id and api_booking.status='completed' limit 20 ; 
          install_ts           | id 
-------------------------------+----
 2016-09-15 09:53:42.511367+00 |  9
 2016-10-12 11:37:11.438715+00 |  9
 2016-10-21 08:55:49.57813+00  |  9
 2017-02-27 06:12:17.362996+00 |  9
 2017-02-27 06:24:59.316051+00 |  9
 2017-02-28 06:15:35.00841+00  |  9
 2017-02-28 06:34:57.766365+00 |  9
 2017-05-23 14:40:54.831525+00 | 14
 2017-06-05 07:47:39.78306+00  | 17
 2017-06-05 07:55:30.171103+00 | 17
 2016-12-06 06:47:43.860581+00 | 19
 2016-09-09 07:34:58.40589+00  | 20
 2016-11-16 09:09:24.466439+00 | 24
 2016-10-03 02:52:24.419793+00 | 24
 2017-05-02 03:48:02.843209+00 | 24
 2017-05-08 10:01:45.77093+00  | 24
 2016-09-09 12:08:27.503695+00 | 27
 2016-09-10 10:05:44.617737+00 | 27
 2016-09-11 10:08:22.791411+00 | 27
 2016-09-12 08:56:31.462979+00 | 27
(20 rows)

现在，我只想要每个user_id的第一行.

Now, I only want the first row from each user_id.

喜欢，

          install_ts           | id 
-------------------------------+----
 2016-09-15 09:53:42.511367+00 |  9
 2017-05-23 14:40:54.831525+00 | 14
 2017-06-05 07:47:39.78306+00  | 17
 2016-12-06 06:47:43.860581+00 | 19
 2016-09-09 07:34:58.40589+00  | 20
 2016-11-16 09:09:24.466439+00 | 24
 2016-09-09 12:08:27.503695+00 | 27

我应该对此使用什么查询?

What query should I use for this ?

推荐答案

这应该可以解决问题:

select distinct on (api_user.id) api_user.id,api_booking.install_ts 
from api_booking inner 
join api_user on api_booking.user_id=api_user.id 
and api_booking.status='completed' 
order by api_user.id,api_booking.install_ts
limit 20 ;

这篇关于在postgres查询中获取第一行的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！