在postgres中按对选择不同的最后一条消息 [英] Select distinct last messages by pairs in postgres

问题描述

我有一个带有消息的表：

 从消息
   * 
 p> 
 
 
这就是我得到的：
 
 
 
  
 
 
 
我确实得到了两个用户之间的唯一对话行，但不是最后一个。
 
 
 
 如何只选择两个用户之间的最后一个唯一对话？ 
 
 
 
已编辑：预期输出：
 
 
 
  
解决方案
这是我的操作方式：
  SELECT DISTINCT ON（user_from，user_to）* 
 FRO M条消息
 ORDER BY user_from，user_to，date + time DESC 
  
 
I have a table with messages:



I need to only one LAST by date and time conversation between two users and here is what i do:
select  distinct on (user_from, user_to)
*
from messages
Here is what I get:



I do get one unique conversations lines between two users, but its not the last one.

How should I select only the last one unique conversation between two users?

EDITED: expected output:

解决方案
Here is how I did it:
SELECT DISTINCT ON (user_from, user_to)   *
FROM messages
ORDER BY user_from, user_to, date+time DESC


                        
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