I’m trying to write a simple login script but I’m getting the resource id #3 error.
code is

Code:

<?php
$host="localhost";
$user="root";
$password="";
$db="admin";
$tableadmin="admin1";

$adminuser= $_POST["adminuser"];
$adminpass= $_POST["adminpass"];

$link = mysql_connect($host,$user,$password)
or die ("couldn't connect to server");
$db = mysql_select_db($db,$link)
or die ("Couldn't select database");

 $result = "SELECT * FROM admin1 WHERE adminuser='$adminuser'  AND adminpass='$adminpass'";   
    $query = mysql_query($result)
    or die ("Couldn't execute query."); 
    echo "<br><br>$query<br><br>";
    //heres the problem
if (mysql_num_rows($result) > 0)   
{
$_SESSION["authenticatedUser"]= $adminuser;
header("Location: loggedon.php");
}
else
{
$_SESSION["message"]="could not connect to database as  $adminuser";
header("Location: loginform.php");
}
?>

Hope someone can help

why wouldnt the SQL statement be called $query and
 the mysql_query be stored as the result?

try this

$query = "SELECT COUNT(*) as cnt FROM admin1 WHERE adminuser='$adminuser'  AND adminpass='$adminpass'";   

$result = mysql_query($query) or die ("Couldn't execute query."); 

$row = mysql_fetch_array($result);

if ($row['cnt'] > 0)   
{
$_SESSION["authenticatedUser"]= $adminuser;
header("Location: loggedon.php");
}

thanks for your help but Still not working

putt @ before mysql_query like this

Code:

$result = @mysql_query($query) or die ("Couldn't execute query.");

still have a problem
I think u Check your HTML for blank lines