ORDER BY GROUP BY

Question

我有一個頁面顯示與您聊天的人員列表。目前它正在顯示你一直在聊天的人和OLDEST消息。但是，它應該顯示NEWEST消息。這是查詢：

SELECT * FROM post 
WHERE fk_user_to = '$userid' 
GROUP BY fk_user_from 
ORDER BY datotime DESC 

和表結構

post_id || fk_user_to || fk_user_from || message || datotime 

Answer 1

這工作了：

SELECT * FROM 
    (SELECT * FROM post ORDER BY datotime DESC) as inv 
    WHERE fk_user_to = '$userid' 
    GROUP BY fk_user_from 
    ORDER BY datotime DESC 

Answer 2

它不會真的這樣做。你永遠不能確定GROUP BY選擇哪一行，因此在它上面放置一個ORDER BY將會產生不可預測的結果。您也不能在GROUP BY之前下單。

試試這個：

SELECT *,MAX(datotime) AS dt fk_user_from 
FROM post 
WHERE fk_user_to = '$userid' 
GROUP BY fk_user_from 
ORDER BY dt 

Answer 3

沒有與普通的SQL沒有簡單的解決方案，我認爲。如果你的數據庫系統允許的子查詢，你得做一個子查詢像

SELECT fk_user_from AS id, MAX(dateotime) AS newest_timestamp FROM post WHERE fk_user_to = '$userid' 

，然後再次使用這兩個領域爲重點，從post選擇。

在MySQL中，可能是這樣的：

SELECT * FROM post P1 
JOIN (
    SELECT fk_user_from, MAX(dateotime) AS newest_dateotime 
    FROM post 
    WHERE fk_user_to = '$userid' 
    GROUP BY fk_user_from 
) AS P2 
ON P1.fk_user_from=P2.fk_user_from 
AND P1.datotime =P2.newest_dateotime 
WHERE P1.fk_user_to = '$userid'