mysql的獲取在小組最後一項通過

Question

嘿，這是我的MySQL查詢

$userid="id of logged in user"; 

---

SELECT id, MAX(post_time) , message, user_name, receiver_id, sender_id 
FROM chat 
WHERE receiver_id =$userid 
OR sender_id =$userid 
GROUP BY GREATEST(receiver_id, sender_id) , LEAST(receiver_id, sender_id) 
ORDER BY id DESC 
LIMIT 0 , 1000 

它給記錄兩個用戶之間不是最後的第一條消息。正如我想要的最後一條消息顯示在郵件收件箱中。我用這個查詢，以顯示在收件箱中唯一的對話與不同用戶的登錄用戶（或U可以說，特定的用戶） 您可以查看演示@http://www.funnenjoy.com/home

Answer 1

我得到這個代碼由我解決了.............如果任何一個需要它幫助了別人也三江源

SELECT m.* 
FROM (
SELECT sender_id + receiver_id -1 AS other_recipient_id, MAX(id) AS id 
FROM chat 
WHERE (sender_id =*user's_id* 
OR receiver_id =*user's_id*) 
GROUP BY sender_id + receiver_id -1 
)mm 
INNER JOIN chat m ON mm.id = m.id 
ORDER BY id ASC 

Answer 2

試試這個，

SELECT id, message, user_name, receiver_id, sender_id 
FROM chat 
WHERE post_time = MAX(post_time) AND (receiver_id =$userid 
OR sender_id =$userid) 
GROUP BY GREATEST(receiver_id, sender_id) , LEAST(receiver_id, sender_id) 
ORDER BY id DESC 
LIMIT 0 , 1000 

Answer 3

這可能有助於你

SELECT id, message, user_name, receiver_id, sender_id 
FROM chat 
WHERE receiver_id =$userid 
OR sender_id =$userid 
HAVING post_time = MAX(post_time) 
GROUP BY GREATEST(receiver_id, sender_id) ,   LEAST(receiver_id, sender_id) 
ORDER BY id DESC 
LIMIT 0 , 1000