$check = mysql_query("
SELECT username
FROM users
WHERE username = '$usercheck'
");
以上是我的代碼,我的問題是我無法創建表stonecrg_password.users
因爲期限。mySQL php,代碼衝突
如何更改上面的代碼,以便它只顯示stonecrg_users
。即使我將代碼更改爲:
$check = mysql_query("
SELECT username
FROM stonecrg_users
WHERE username = '$usercheck'
");
它仍然會顯示stonecrg_password.stonecrg_users
。
如果我理解你,我會立即死:P – hungneox 2011-12-18 09:26:14
是'stonecrg_password'數據庫,或者它應該是在數據庫中的表? – 2011-12-18 09:29:54
Dreamweaver與此有何關係? – 2011-12-18 09:38:31