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我做這個請求: $uid = db_query("SELECT uid FROM tablename WHERE mail = '[email protected]' LIMIT 1")
SQL請求返回源ID
並獲得儘可能結果:
資源ID#310
但我想要得到的UID。錯誤在哪裏?
我做這個請求: $uid = db_query("SELECT uid FROM tablename WHERE mail = '[email protected]' LIMIT 1")
SQL請求返回源ID
並獲得儘可能結果:
資源ID#310
但我想要得到的UID。錯誤在哪裏?
db_query("SELECT uid FROM tablename WHERE mail = '[email protected]' LIMIT 1");
只是給你一個「答案」,但你想要它的價值。試試這個:
$result = db_query("SELECT uid FROM tablename WHERE mail = '[email protected]' LIMIT 1");
$row = db_fetch_array($result);
$uid = $row[0];
很酷。非常感謝您的回答。我剛剛開了一個新帖子,也許你也可以在這裏看到它? – Chris 2010-11-26 23:47:11