5

これは私のfqlです:

$fql = "SELECT uid, name, online_presence, status 
         FROM user 
        WHERE uid IN ( SELECT uid2 FROM friend WHERE uid1 = '".$user_id."') ";

値はonline_presence常に を返しnullます。

4

2 に答える 2

0

クエリにエラーがあります...

交換するだけ

WHERE uid IN

WHERE 2 IN

クエリ

$fql = "SELECT uid, name, online_presence, status 
         FROM user 
        WHERE 2 IN ( SELECT uid2 FROM friend WHERE uid1 = '".$user_id."') ";

SQLFiddle デモ

于 2013-04-08T20:10:19.140 に答える