elseif(isset($_POST['submit']))
{
// Look for their user
$lookuser = mysql_query("SELECT * FROM `users` WHERE username='". mysql_escape_string($_POST['username']) ."'");
// If we find a row
if(mysql_num_rows($lookuser) > 0)
但我其他爲,回聲:An error has occured. <br> If you are sure you entered your username correctly, please contact an administrator.
mysql_num_rows PHP
我試圖呼應$_POST['username'];
都工作地很好。我確定我的用戶存在,這工作正常。
PHP的錯誤,我得到: Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in password.php on line 23
您在查詢中沒有執行任何錯誤檢查。爲什麼不?如果查詢失敗,這也就不足爲奇了。此參考問題顯示如何:http://stackoverflow.com/questions/6198104/reference-what-is-a-perfect-code-sample-using-the-mysql-extension – 2012-03-19 22:16:21
也許多一點的代碼會有所幫助。 – 2012-03-19 22:16:35
你是否回顯了mysql_error()? $ lookuser = mysql_query(「SELECT * FROM'users' WHERE username ='」。mysql_escape_string($ _ POST ['username'])。「'」)or die(mysql_error()); – 2012-03-19 22:17:17