选择最后一对不同的配对 - Select last distinct pair

Select last distinct pair

本文关键字：选择最后 | 更新日期: 2023-09-27

好的，我们有收件箱表，用于保存用户相互发送的消息。这是表格：

CREATE TABLE IF NOT EXISTS `inbox` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`fromid` int(10) unsigned NOT NULL DEFAULT '0',
`toid` int(10) DEFAULT NULL,
`message` text CHARACTER SET utf8 NOT NULL,
`time` datetime NOT NULL DEFAULT '0000-00-00 00:00:00',
PRIMARY KEY (`id`),
KEY `toid` (`toid`),
KEY `fromid` (`fromid`),
KEY `fromid_2` (`fromid`,`toid`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1  ;

fromid和toid是用户的id。我们有他们的id，消息发送的时间。我们需要的是一个查询，它将返回"我们的用户"（管理员）未回复的所有消息。表帐户跟踪用户。简化：

CREATE TABLE IF NOT EXISTS `accounts` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`our` int(1) unsigned NOT NULL DEFAULT '0',
PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1;

因此，基本上，我们需要一个查询，向我们提供管理员（我们的用户）未回复消息的用户、他们的计数以及他们发送给ADMIN的最后一条消息的日期，从最后一条到最老一条排序。

到目前为止，我们只有一些基本的问题，我们没有提出任何合理的问题，我可以发布。

提前谢谢。

编辑：据我所见，我们首先需要在收件箱表中找到两个DISTINCT用户的最后一次互动。。。然后检查&只过滤那些发送给我们的用户

这个怎么样？

SELECT i.* FROM inbox as i 
WHERE (i.toid, i.fromid) NOT IN 
(SELECT i2.fromid, i2.toid FROM inbox as i2 WHERE i2.`time` >= i1.`time` AND i2.id = 1);

使用join:的另一种方法

SELECT DISTINCT i1.* 
FROM inbox as i1 LEFT JOIN inbox as i2 
    ON  i1.toid = 1 AND 
        i1.fromid = i2.toid AND 
        i1.toid = i2.fromid AND 
        i1.`time` <= i2.`time`
WHERE i2.id IS NULL;

下面给出了两种可能的解决方案：LEFT JOIN解决方案应该性能更好。

LEFT JOIN解决方案

SELECT 
 i.fromid, COUNT(*) AS unread, MAX(i.time) AS lastmsg 
FROM inbox AS i 
INNER JOIN accounts AS a 
 ON i.toid = a.id 
LEFT JOIN inbox AS i2 
 ON i.fromid = i2.toid AND i.toid = i2.fromid AND i.time <= i2.time 
WHERE a.our = 1 AND i2.id IS NULL
GROUP BY i.fromid
ORDER BY lastmsg DESC;

不在解决方案

SELECT 
 i.fromid, COUNT(*) AS unread, MAX(i.time) AS lastmsg
FROM inbox AS i 
INNER JOIN accounts AS a ON i.toid = a.id 
WHERE a.our = 1 AND 
 (i.toid, i.fromid) 
 NOT IN (SELECT i2.fromid, i2.toid FROM inbox AS i2 WHERE i2.time >= i.time) 
GROUP BY i.fromid
ORDER BY lastmsg DESC;