Question

CREATE TABLE `messages` (
  `message_id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `message_project_id` int(7) unsigned NOT NULL DEFAULT '0',
  `message_time` int(11) unsigned NOT NULL DEFAULT '0',
  `message_from_user_id` int(7) unsigned NOT NULL DEFAULT '0',
  `message_to_user_id` int(7) unsigned NOT NULL DEFAULT '0',
  `message_details` text COLLATE utf8_unicode_ci NOT NULL,
  PRIMARY KEY (`message_id`)
)


CREATE TABLE `project` (
  `project_id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `project_user_id` int(7) unsigned NOT NULL DEFAULT '0',
  `project_name` varchar(255) COLLATE utf8_unicode_ci NOT NULL DEFAULT '',
  `project_status` varchar(50) COLLATE utf8_unicode_ci NOT NULL DEFAULT '',
 PRIMARY KEY (`project_id`)
)

我希望retreive是每个项目的用户＃2的最新消息。

用户＃2是项目的所有者，因此可以接收来自许多相关方的消息。

实际页面将显示用户＃2的“待办事项”列表。我想找到当前打开的项目的任何消息，其中用户＃2已收到消息，但尚未发送一个

因此，如果用户＃1向用户＃2发送了一条消息，则消息表中会有一行

message_id | project_id | message_time | message_from_user_id | message_to_user_id | message_details
30 | 12 | 1304707966 | 1 | 2 | Hello user number two, thank you for your interest in my project
31 | 12 | 1304707970 | 2 | 1 | Hello user number one, Your project looks interesting
32 | 12 | 1304707975 | 3 | 1 | Hello user number one, here is my first message, im user number three.  I want to do your project
32 | 13 | 1304707975 | 7 | 1 | Hello user number one, here is my first message, im user number seven.  I want to do your other project

到目前为止我尝试了但是没有完成工作： //这将为我提供每个项目的最新消息，但不会被用户分开

SELECT cur.*, p.*
FROM messages cur
LEFT JOIN messages next ON cur.message_project_id = next.message_project_id AND cur.message_time < next.message_time
LEFT JOIN project p ON p.project_id = cur.message_project_id
WHERE next.message_time IS NULL
AND (cur.message_from_user_id = 2 OR cur.message_to_user_id = 2)
AND (p.project_status LIKE 'open' OR p.project_status LIKE 'started')
AND p.project_user_id = 2
ORDER BY cur.message_time DESC

//这将按用户分隔，但不会返回最新的消息文本

SELECT *
FROM messages m
LEFT JOIN project p ON p.project_id = m.message_project_id
WHERE (message_from_user_id = 2 OR message_to_user_id = 2 )
AND (p.project_status LIKE 'open' OR p.project_status LIKE 'started')
AND p.project_user_id = 2
GROUP BY project_id
ORDER BY message_time DESC

一旦数据返回Php i，检查最近的消息是否是TO用户＃2，如果是，则将“你需要回复”消息发布到他的屏幕。

Answer 1

如果您提供了一些涵盖大部分案例的示例输出，那可能会有所帮助。从您的评论来看，经过多次重读后，我认为这就是您所寻找的：

SELECT
    <column list>
FROM
    Messages M
INNER JOIN Projects P ON
    P.project_id = M.message_project_id AND
    P.project_status IN ('open', 'started') AND
    P.project_user_id = 2
WHERE
    M.message_to_user_id = 2 AND
    NOT EXISTS (
        SELECT *
        FROM Messages M2
        WHERE
            M2.message_from_user_id = 2 AND
            M2.message_project_id = M.message_project_id AND
            M2.message_to_user_id = M.message_from_user_id AND
            M2.message_time >= M.message_time
    )

这将为用户提供所有项目的所有消息，他们尚未将消息发送回该项目的发件人。您当然可以添加ORDER BY。

Answer 2

这是否有效：


SELECT *
FROM messages m
LEFT JOIN project p ON (p.project_id = m.message_project_id)
WHERE (m.message_from_user_id = 211 OR m.message_to_user_id = 211 )
AND p.project_status IN('open','started') AND p.project_user_id = 2
GROUP BY p.project_id
ORDER BY m.message_time DESC

希望有所帮助

需要一个mysql查询来解决“group by”和“order by”不返回最新消息

2 个答案: