1

使用以下查询，我正在寻找一种解决方案来获取具有某些条件的最新记录。但它给了我第一个记录，而不是最新的。我认为它只考虑组请告诉我

SELECT * FROM `contacts` WHERE `id_receiver`=1  GROUP BY `id_thread` ORDER BY created DESC

id   id_sender   id_thread sender_email   id_receiver      created(datetime)
1    2             2         51             1                2012-03-24 13:44:48
2    4             4         1              5                 2012-04-26 13:46:05
3    2             2         51             1                2012-04-09 12:12:30

所需输出

id   id_sender  id_thread sender_email   id_receiver      created(datetime)
3    2           2         51             1                2012-04-09 12:12:30

我做了一个测试，只是交换了 order by 和 group by ，给了我一个错误。

任何人都可以看看这个？谢谢。

编辑编辑的问题，忘记写 id_thread

4

2 回答 2

3

GROUP BY id_thread当你的表中没有id_thread列时你怎么能？

SELECT * 
FROM contacts 
WHERE id_receiver = 1 
                                 --- GROUP BY id_thread
                                 --- removed 
ORDER BY created DESC
LIMIT 1                          --- only show one row

根据您的评论，您想要的是createdevery 的最新（按 排序）行id_thread，这是一个不同且更复杂的查询。这种查询甚至还有一个标签，名为[greatest-n-per-group].

SELECT c.* 
FROM contacts AS c
  JOIN
    ( SELECT id_thread, MAX(created) AS created 
      FROM contacts 
      WHERE id_receiver = 1  
      GROUP BY id_thread
    ) AS g 
    ON (g.id_thread, g.created) = (c.id_thread, c.created) 
WHERE c.id_receiver = 1
于 2012-04-09T07:39:00.880 回答
0

如果记录是按顺序排列的，那么您也可能按 id 排序——当且仅当它是按顺序创建的——

于 2012-04-09T07:26:06.140 回答