Php Login Script

[loonie35]

whats the problem?[error]Warning: mysql_result(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\sludinajumi\admin\index.php on line 10[/error]

<?phprequire("../include/config.php");if(isset($_POST['user'])){	$user = quote_smart($_POST['user']);	$password = quote_smart(md5($_POST['password']));		$result = mysql_query("SELECT COUNT(*) FROM users_yoy WHERE user = $user AND password = $password");	if(mysql_result($result,0,'COUNT(*)') > 0)	{		$_SESSION['ieligojies'] = 1;	}}?><form method="post" action=""><table><tr><td>User:</td><td><input type="text" name="user"></td></tr><tr><td>Password:</td><td><input type="password" name="password"></td></tr><tr><td></td><td><input type="submit" value="Ieiet"></td></tr></table></form>

in db all is right.

[jeffman]

mysql_query() is returning false, which you should be checking before doing anything with its return result.So why is it returning false? A possibility: do you have a quote_smart() function in your include file or anywhere else attached to this code? It's not built-in, but a lot of samples out there refer to such a function. You just need to have one.But why would you quote_smart an MD5 return value anyway? Or a username?It's also possible mysql_query() produced no results because the username and/or password were incorrect.

[deboni]

have you tried this way?

$result = mysql_query("SELECT COUNT(*) FROM users_yoy WHERE user = '$user' AND password = password('$password')");

just in case the password (as it's supposed) was entered as hashed password.regardsregards