Где ошибка синтаксиса в запросе mysql

Здравствуйте друзья. Пишу так

<?php 
$id=$_GET['id'] ;
if (isset($_POST['edit_slide'])) 
                            {
                                $title = clean($_POST ['title']) ;
                                $desc = clean($_POST ['desc']) ;
                                $title_eng = clean($_POST ['title_eng']) ;
                                $desc_eng = clean($_POST ['desc_eng']) ;                                    
                                $file=$_FILES['image']['tmp_name'];
$random_date = date('Y-m-d') ;                              
$random = (rand(0,9999));
$mic = uniqid();
$image= file_get_contents($_FILES['image']['tmp_name']);
$image_name= ($random_date.$mic.$random.$_FILES['image']['name']);

        move_uploaded_file($_FILES["image"]["tmp_name"],"../assets/images/" . $image_name);
        $location="../assets/images/" . $image_name;
        $query =("UPDATE sliders SET title='$title', desc='$desc', 
  location='$location' WHERE id='$id'");
  $result3 = mysqli_query($con , $query ) or die("Ошибка " . 
 mysqli_error($con)); 
        if($result3)
{
mysqli_query ($con ,"UPDATE eng_sliders SET title='$title_eng', 
desc='$desc_eng', location='$location' WHERE id='$id'") or die("Ошибка " . 
mysqli_error($con));
    echo "<meta http-equiv='refresh' content='0; URL=index.php>";
     }
                            }
 ?>

Выдаёт ошибку Ошибка You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'desc='Содержание поля', location='../assets/image' at line 1 Где ошибка помогите пожалуйста.

PS Искренне прошу прощения за го**нокод