Why Not it doesn't Work???

[Kasya]

<? Error_Reporting(E_ALL & ~E_NOTICE);

function auth($user_name,$user_pass) {
mysql_connect("localhost","root","") or die("ERROR:1");
mysql_select_db("Kasya") or die("ERROR:2");
$query = "SELECT * FROM `user_info` VALUES('user_name','user_pass')";
$result = mysql_query($query);

$inf = mysql_fetch_array($result);

if($inf['user_name'] == $user_name and $inf['user_pass'] == $user_pass) {


echo("Access Allow!");


}

else {
die("Access Denied! Invalid Username or Password!");

}

}



echo "

<form action='$PHP_SELF' method='post'>
UserName: <input type='text' name='name'><br/>
Password: <input type='password' name='pass'><br/>
<input type='submit' name='ok' value='Enter'>
</form>

";



$user = $_POST['user'];
$pass = $_POST['pass'];
$ok = $_POST['ok'];

auth($user,$pass);

?>

Errors:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in w:\home\localhost\www\kasya\db.php on line 9

[iam_clint]

Thats saying that $result is empty more then likely your sql query is incorrect.

[Banfa]

The query

$query = "SELECT * FROM `user_info` VALUES('user_name','user_pass')";

Is wrong there is no VALUES clause to a select query. I think you mean either

$query = "SELECT * FROM `user_info`";

or

$query = "SELECT user_name,user_pass FROM `user_info`";