Why Not it doesn't Work???

[Kasya]

<? Error_Reporting (E_ALL & ~E_NOTICE);

function auth($user_name ,$user_pass) {
mysql_connect(" localhost","roo t","") or die("ERROR:1");
mysql_select_db ("Kasya") or die("ERROR:2");
$query = "SELECT * FROM `user_info` VALUES('user_na me','user_pass' )";
$result = mysql_query($qu ery);

$inf = mysql_fetch_arr ay($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_SE LF' 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($user,$pas s);

?>

Errors:

Warning: mysql_fetch_arr ay(): supplied argument is not a valid MySQL result resource in w:\home\localho st\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_na me','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`";