Question

我有三张桌子，我只会在这里向您展示相关栏目

表：组列：group_id，name。

表：groups_to_message列：   group_id，message_id

表：消息列：message_id，   创建（日期）

我需要基本上找到每个组的最后一条消息，而不显示重复的组。

我尝试过像这样使用group：

SELECT m.created, g.group_id 
FROM groupss as g 
JOIN group_to_message as gm ON (g.group_id = gm.group_id) 
JOIN messages as m 
GROUP BY g.group_id 
ORDER BY m.created DESC

这会导致成功分组，但是在ORDER BY之前完成，因此在排序之前会先获得第一个结果。

任何帮助表示感谢。

Answer 1

如果我正确地阅读你的问题，请使用MAX（）。这应该可以获得每个组的最新项目

示例...

SELECT MAX(m.created), g.group_id 
FROM groupss as g 
JOIN group_to_message as gm ON (g.group_id = gm.group_id) 
JOIN messages as m ON (gm.message_id = gm.message_id)
GROUP BY g.group_id 
ORDER BY m.created DESC

Answer 2

SELECT  m.created, g.group_id
FROM    groupss g
JOIN    messages m
ON      m.id = 
        (
        SELECT  mi.id
        FROM    group_to_message gm
        JOIN    message mi
        ON      mi.message_id = gm.message_id
        WHERE   gm.group_id = g.group_id
        ORDER BY
                gm.group_id DESC, mi.created DESC
        LIMIT 1
        )

Answer 3

这是未经测试的，但它应该有效：

SELECT m.created, g.group_id 
FROM groups as g 
JOIN group_to_message as gm ON (g.group_id = gm.group_id) 
JOIN messages as m ON (gm.message_id = m.id)
GROUP BY g.group_id HAVING m.created = MAX(m.created)

mysql得到的顺序发生在group by之前

3 个答案: