PostgreSQL 通过 GROUP BY 删除重复项

Question

我想打印一个人的最后一条消息，但每个人只应打印他的最新消息。我使用 PostgreSQL 10。

+-----------+----------+--------------+
| name      |   body   |  created_at  |
+-----------+----------+--------------+
| Maria     | Test3    |  2017-07-07  |
| Paul      | Test5    |  2017-06-01  |
+-----------+----------+--------------+

我已经用下面的 SQL 查询尝试过，这给了我确切的结果，但不幸的是，人们在其中加倍了。

+-----------+----------+--------------+
| name      |   body   |  created_at  |
+-----------+----------+--------------+
| Maria     | Test3    |  2017-07-07  |
| Paul      | Test5    |  2017-06-01  |
+-----------+----------+--------------+

+-----------+----------+--------------+
| name      |   body   |  created_at  |
+-----------+----------+--------------+
| Maria     | Test1    |  2016-06-01  |
| Maria     | Test2    |  2016-11-01  |
| Maria     | Test3    |  2017-07-07  |
| Paul      | Test4    |  2017-01-01  |
| Paul      | Test5    |  2017-06-01  |
+-----------+----------+--------------+

我尝试使用 DISTINCT 删除重复项，但不幸的是我收到此错误消息：

SELECT * FROM messages 
WHERE receive = 't'
GROUP BY name
ORDER BY MAX(created_at) DESC

ERROR: SELECT DISTINCT ON expressions must match initial ORDER BY expressions LINE 1: SELECT DISTINCT ON (name) * FROM messages ^ : SELECT DISTINCT ON (name) * FROM messages WHERE receive = 't' GROUP BY name ORDER BY MAX(created_at) DESC

您对我如何解决这个问题有什么想法吗？

Answer 1

您将使用DISTINCT ON如下：

SELECT DISTINCT ON (name) * 
FROM messages 
WHERE receive = 't'
ORDER BY name, created_at DESC

那是：

• GROUP BY不需要任何条款

• 中列出的列DISTINCT ON(...)必须出现在ORDER BY子句的前面

• ...后面是应该用来打破组的列（这里是created_at）

请注意，查询的结果distinct on始终按子句中的列排序（因为这种排序用于标识应保留哪些行）。

如果您想更好地控制排序顺序，则可以使用窗口函数：

SELECT *
FROM (
    SELECT m.*, ROW_NUMBER() OVER(PARTITION BY name ORDER BY created_at DESC) rn
    FROM messages m
    WHERE receive = 't'
) t
WHERE rn = 1
ORDER BY created_at DESC