My code:
$fileid = $_GET['imgid'];
$fileid = (int)$fileid; //id is int type in photos table
require 'database.php';
//get the image sourc name
$q = "SELECT src form photos WHERE id='$fileid'";
$result = $mysqli->query($q) or die(mysqli_error($mysqli));
if ($result) 
{
    $row = $result->fetch_object();
    $filename = $row->src;
ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'photos WHERE id='12'' at line 1
 
     
     
    