【发布时间】:2014-12-16 16:46:18
【问题描述】:
我认为这很容易找到解决方案,但我找不到与我正在尝试做的事情相匹配的问题。
这是我想看到的(这个查询不起作用):
SELECT DISTINCT ON (conversation_id)
*
FROM messages
ORDER BY created_at DESC
换句话说,返回具有最新 created_at 日期的完整行,其中 conversation_id 不同,然后按 created_at DESC 排序。
我在下面写了一个查询,它完全符合我的要求,但我认为我过于复杂了。我还假设最近的记录将具有最高的 ID,并在子查询中按其排序,而不是“created_at”(它返回与 created_at DESC 相同的顺序,但对读者不太友好)。
SELECT
m.*
FROM (
SELECT
DISTINCT ON (conversation_id) conversation_id,
id
FROM messages
ORDER BY conversation_id, id DESC
) as t
INNER JOIN messages as m
ON t.id = m.id
ORDER BY t.id DESC
在制作这个查询的更简单版本方面的任何帮助都会很棒。
为澄清添加了示例
如果消息表如下所示:
id, conversation_id, created_at, subject
1, 2, "2014-10-21 00:01:00", "subject 1"
2, 43, "2014-10-21 00:02:00", "subject 2"
3, 12, "2014-10-21 00:03:00", "subject 3"
4, 2, "2014-10-21 00:04:00", "subject 4"
5, 43, "2014-10-21 00:05:00", "subject 5"
查询应返回以下内容:
id, conversation_id, created_at, subject
5, 43, "2014-10-21 00:05:00", "subject 5"
4, 2, "2014-10-21 00:04:00", "subject 4"
3, 12, "2014-10-21 00:03:00", "subject 3"
【问题讨论】:
标签: sql postgresql greatest-n-per-group