Question

mysql_fetch_array not working in my code :- I got an error like this …

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\xampp\htdocs\finalreports\generatereport.php on line 38

My code is So far ….

if(array_key_exists('server_guid',$_GET))
{
    $guids = $_GET['server_guid'];
    $guid_array = explode(",",$guids);

    //$reporttype = "Server Resources";

    for($i=0 ; $i<count($guid_array); $i++)
    {
        $query = "select vid from vendor_registration where bussname='".$guid_array[$i]."'";
        $result = mysql_query($query);



        while($row = mysql_fetch_array($result))
        {
            $name_array[$i] = $row[0];

        }
    }
}

Answer 1

Make sure the result is not tainted:

$result = mysql_query($query) or die(mysql_error());