我製作了一個使用PHP和mysql的小型郵件系統。我在消息表中將用戶鏈接爲sender
和receiver
。我可以列出2個用戶之間的對話。但我無法找到顯示每個對話1個消息結果的方式!消息系統,顯示對話列表
說我有用戶1,2和3。所以我的信息表可以是這樣的:
===================================================================
ID || sender || receiver || text || date
===================================================================
1 || 1 || 2 || Hello! || timestamp stuff
2 || 2 || 1 || Hi! ||
3 || 1 || 3 || Boo! ||
所以談話被鏈接爲
SELECT * FROM messages WHERE (sender=1 OR sender=2) AND (receiver=1 OR receiver=2);
這樣會得到我的2個用戶之間的完整對話(真正簡化)。
但是,如何從每次對話中獲得1條消息?
我已經試過這樣的事情:
(SELECT m.text,m.date,sender.username,receiver.username
FROM messages AS m
INNER JOIN users AS sender on m.sender=user.ID
INNER JOIN users AS receiver ON receiver.ID=m.receiver
WHERE m.sender=1 OR m.receiver=1
ORDER BY messages.date DESC LIMIT 1);
但顯然這並沒有做我想做的,因爲只有返回一行。我想我知道我需要做什麼,我只是不知道該怎麼做? 我想我需要這樣的東西:
SELECT sender.username,receiver.username,m.text,m.date FROM messages AS m
INNER JOIN users AS sender ON sender.ID=m.sender
INNER JOIN users AS receiver ON receiver.ID=m.receiver
LEFT JOIN (SELECT mes.ID
FROM messages AS mes
WHERE mes.sender=1 OR mes.receiver=1
ORDER BY mes.date DESC LIMIT 1) AS convo
ON convo.ID=m.ID;
會這樣的工作?
如果你想看到我的完整的SQL:
(SELECT sender.ID AS senderID,sender.username AS senderUsername,sender.firstname AS senderFirstname,sender.lastname AS senderLastname,sender.othername AS senderOthername,sender.url_name AS senderUrl_name,senderI.path as sender_default_image,
receiver.ID AS receiverID,receiver.username AS receiverUsername,receiver.firstname AS receiverFirstname,receiver.lastname AS receiverLastname,receiver.othername AS reciverOthername,receiver.url_name AS recieverUrl_name,receiverI.path as reciever_default_image,
messages.message,messages.date,messages.read
FROM '.$wpdb->prefix.'fishy_messages AS messages
INNER JOIN '.$wpdb->prefix.'fishy_users AS sender ON sender.ID=messages.sender
LEFT JOIN '.$wpdb->prefix.'fishy_profile_default_image AS senderDP ON senderDP.userID=sender.ID
INNER JOIN '.$wpdb->prefix.'fishy_users AS receiver ON receiver.ID=messages.receiver
LEFT JOIN '.$wpdb->prefix.'fishy_profile_default_image AS receiverDP ON receiverDP.userID=receiver.ID
LEFT JOIN '.$wpdb->prefix.'fishy_images AS senderI ON senderI.ID=senderDP.imageID
LEFT JOIN '.$wpdb->prefix.'fishy_images AS receiverI ON receiverI.ID=receiverDP.imageID
WHERE messages.sender=%d OR messages.receiver=%d ORDER BY messages.date DESC LIMIT 1)
是的,我正在使用WordPress的:P
我把它放在內部查詢中,因爲我想從每個對話中有一個對話,其中有一個對話。從我的例子說我應該得到2個結果,其中sender.ID = 1或receiver.ID = 1。所以外部查詢的限制只會返回一個會話不是嗎?如果我搜索user.ID = 2,我應該只能得到1個結果。謝謝回答! – sourRaspberri
那就對了!我編輯了查詢..請檢查 –
差不多!你也許是對的。這只是我的一個愚蠢的XD感謝您的幫助。當我知道了,我會讓你知道並接受你的答案:p – sourRaspberri