[Solved] Invalid argument supplied for foreach with code

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Lots of questions regarding this sort of error. First its a warning not an error so your code will continue to work just fine. You can suppress by using error_reporting(E_ERROR) which will only show errors that are fatal. Otherwise in your code if you dont want to see this error you need to do something like..

//only try loop if there is an array
if(is_array($_SESSION["qty"])){ 
  foreach ($_SESSION["qty"] as $each_item) { 
    $item_id = $each_item['qty'];

    $update_qty = mysqli_query("update cart set qty='$qty' LIMIT 1");
    //change to mysqli_fetch_array!!
    while ($row = mysqli_fetch_array($update_qty)) {
        //you will only get one row in here like this
        $_SESSION['qty']=$qty; //maybe do $_SESSION['qty'][]=$qty;
    }
    $total = $total * $each_item['qty'];
  }
}else{
   echo 'No array to be found here';
}

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solved Invalid argument supplied for foreach with code