Question

我正在构建一个聊天应用程序，我希望列出最近发送消息的用户及其id, name, pic, and last message (id, timestamp, content)

我的两张桌子看起来像这样：

users:
------
id
name
pic

messages:
---------
id
content
timestamp
user_id (foreign key of user's id)

以下是我尝试的查询：

SELECT U.*, 
    MAX(M.id) as lastmessage_id 
FROM users U, messages M 
WHERE M.user_id=U.id 
GROUP BY user_id 
ORDER BY lastmessage_id DESC

它会列出最近收到消息的用户，但不包含最后一封消息的“内容”

当前输出：

id, name, pic, lastmessage_id

必需的输出：

id, name, pic, lastmessage_id, lastmessage_content, lastmessage_timestamp

Answer 1

首先找到每个用户的所有最新消息ID，然后将其与消息表连接以获取相应的行：

select u.*,
    m.*
from users u
join messages m on u.id = m.user_id
join (
    select user_id,
        max(id) as id
    from messages
    group by user_id
    ) t on m.id = t.id
        and m.user_id = t.user_id
order by m.id desc;

此外，始终使用显式现代连接语法而不是旧的基于逗号的语法。

Answer 2

func gotoNextVC() {
    let destinationVC = YourViewController(nibName: "YourViewController", bundle: nil)
    destinationVC.userAcount = self.userAcount
    self.navigationController?.pushViewController(destinationVC, animated: true)

    // OR
    self.present(destinationVC, animated: true, completion: nil)
}

Answer 3

试试这个......

select t1.id,t1.name,t1.pic,t1.lastmessage_id from (SELECT U.id,U.name,U.pic,(M.id) as lastmessage_id FROM users U, messages M WHERE M.user_id=U.id ORDER BY lastmessage_id DESC) t1
GROUP BY t1.id

Answer 4

在脚本中，显示users表中的所有列，但只显示messages表中的id。将脚本更改为低于1。

SELECT U.*, 
    MAX(M.id) as lastmessage_id , M.content, M.timestamp
FROM users U, messages M 
WHERE M.user_id=U.id 
GROUP BY user_id 
ORDER BY lastmessage_id DESC

Answer 5

尝试使用JOIN：

SELECT U.*, MAX(M.id) AS lastmessage_id 
FROM users U

JOIN MESSAGES M ON U.id = M.user_id

GROUP BY user_id 
ORDER BY lastmessage_id DESC

MySQL Query列出最近发送消息的用户

5 个答案: