选择每个会话的最后一条消息

Question

我有这张桌子：

fromJID | toJID | sentDate | body

我需要用户最后五个对话的最后一条消息（toJID）

我试试这个，但这会返回第一个对话的消息，而不是最后一个消息。

SELECT 
    body as mensagem, 
    fromJID, 
    toJID,
    sentDate 
FROM ofmessagearchive 
WHERE toJID = '1945' 
GROUP BY fromJID 
ORDER BY sentDate DESC 
LIMIT 0, 5 

数据示例：

fromJID| toJID  | sentDate      | body (message)
'1945' | '2042' | 1383934233976 | '1\n'
'1945' | '2042' | 1383934234429 | '2\n'
'1945' | '2042' | 1383934234430 | '3\n'
'2042' | '1945' | 1383934237053 | '1\n'
'2042' | '1945' | 1383934237374 | '2\n3'
'2042' | '1945' | 1383934237523 | '\n'
'2042' | '1945' | 1383934242018 | '3\n'
'1945' | '1946' | 1383934364814 | '1\n'
'1945' | '1946' | 1383934365118 | '2\n'
'1945' | '1946' | 1383934365366 | '3\n'
'1946' | '1945' | 1383934367271 | '1\n'
'1946' | '1945' | 1383934367517 | '2\n'
'1946' | '1945' | 1383934367782 | '3\n'

谢谢！

Answer 1

试试这个，使用子查询获取每个对话的最后一个帖子并将其加入主表： -

SELECT a.body as mensagem, 
    a.fromJID, 
    a.toJID,
    a.sentDate
FROM ofmessagearchive a
INNER JOIN
(
    SELECT fromJID, toJID, MAX(sentDate) AS MaxSentDate
    FROM ofmessagearchive
    GROUP BY fromJID, toJID
) b
ON a.fromJID = b.fromJID
AND a.toJID = b.toJID
AND a.sentDate = b.MaxSentDate
WHERE ta.oJID = '1945'
ORDER BY a.sentDate DESC
LIMIT 5