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我正在尝试与相关的最后一条消息进行对话，并按时间排序它们是否已阅读该消息。让我们去展示我的逻辑。

我创建了 3 个表：inbox_join / inbox_msg / users

在第一个表“收件箱加入”上，我有关于谁有积极讨论的数据。在这种情况下，我们有 id_user - "1" 和 id_user_2 - "4" 他们有一个对话。

在 inbox_msg 表中，我有文本消息、将显示消息的 id 对话以及其他易于理解的字段。

收件箱连接表

在此处输入图像描述

收件箱_msg 表 在此处输入图像描述

用户表

在此处输入图像描述

我做了一个可以正常工作的查询，但我的问题是我不能在 inbox_msg 表上找到发生的_at。我想找到一个更好的解决方案来获得我想要的结果，我无法订购我正在寻找的方式。

这是我的查询

SELECT DISTINCT (
inbox_join.id_conversation
), user_chat.name AS name_conv, user_chat.surname AS surname_conv, user_chat.username as username_conv, user_chat.id as id_chat, image_upload.name_image, (

SELECT DISTINCT (
message
)
FROM inbox_msg
WHERE inbox_join.id_conversation = inbox_msg.id_conversation
ORDER BY occured_at DESC 
LIMIT 1
) AS last_msg, users.name, users.surname
FROM inbox_join

INNER JOIN users ON users.id = inbox_join.id_user
INNER JOIN users AS user_chat ON user_chat.id <> 1 AND (inbox_join.id_user_2 = user_chat.id || inbox_join.id_user = user_chat.id)
INNER JOIN image_upload ON image_upload.id_image = user_chat.profile_image
WHERE inbox_join.id_user = 1
OR inbox_join.id_user_2 = 1

想要选择关于用户 1 的对话的结果：

id_conversation | id_user | name | surname | username | last_msg | occured_at_last_msg | read_msg |

       1           4         E        S         E           Yes            1380724676        0
       4           5         G        E         K           Good           1380724675        0
4

2 回答 2

1

询问：

SELECT  im.id_conversation,
        im.id_user,
        u.name,
        u.surname,
        u.username,
        im.message AS last_msg,
        im.occured_at AS occured_at_last_msg,
        im.read_msg 
FROM inbox_msg im
JOIN users u
ON u.id_user = im.id_user
JOIN (SELECT id_conversation,
             MAX(occured_at) AS occured_at
      FROM inbox_msg
      GROUP BY id_conversation) im2
ON im2.id_conversation = im.id_conversation
AND im2.occured_at = im.occured_at
ORDER BY im.occured_at DESC
于 2013-10-16T12:07:28.313 回答
-1

我做了这个查询，应该可以正常工作，我想收到关于这个查询的评论。

SELECT DISTINCT (
            im.id_conversation
            ), users.name, users.surname, users.username, image_upload.name_image, im.message as last_msg, im.occured_at, im.read_msg
            FROM inbox_join
            INNER JOIN (

            SELECT sub . * 
            FROM (

            SELECT DISTINCT (
            id_conversation
            ), id_user, message, occured_at, read_msg
            FROM inbox_msg
            WHERE id_user <> 1
            ORDER BY occured_at DESC
            ) AS sub
            GROUP BY sub.id_user
            ORDER BY sub.occured_at DESC
            ) AS im ON im.id_conversation = im.id_conversation
            INNER JOIN users ON im.id_user = users.id
            INNER JOIN image_upload ON users.profile_image = image_upload.id_image
            WHERE inbox_join.id_user = 1 || inbox_join.id_user_2 = 1
于 2013-10-16T11:45:00.357 回答