Question

* Resolved *

I implemented SimpleCart(js) to my one project.

One can register and login.
Once logged in you can use the cart to purchase items.

I have one problem, because simplecart(js) saves the cart to localstorage, if you login with a other account. You see the same cart results.

Does anyone have a php sql solution for this, to store the cart results into sql for each individual session user?

I resolved this issue with this,
It’s a basic technique, but it works perfectly for what I need, I’ll still tweak the php to check if the cart item exists in the db, whether to update or create.

jQuery

$(".saveCart").click(function(){
        $(".itemRow").each(function(){
            var custId = "<?php echo $_SESSION['Username'] ?>";
            var itemName = $(this).find(".item-name").text();
            var itemPrice = $(this).find(".item-price").text();
            var itemQty = $(this).find(".item-quantity").text();
            var itemTotal = $(this).find(".item-total").text();

            $.ajax({
                // Enter below the full path to your "cart" php file
                url: "cart.php",
                type: "POST",
                data: {custId: custId, itemName: itemName, itemPrice: itemPrice, itemQty: itemQty, itemTotal: itemTotal},
                cache: false,
                success: function (html) {
                  // If form submission is successful
                  if ( html == 1 ) {
                  }
                  // Double check if maths question is correct
                  else {
                  }
                }
            });            

        });
    });

PHP

<?php

$con = mysql_connect("localhost","root","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("test", $con);

$sql="INSERT INTO cart (Id, itemName, itemPrice, itemQty, itemTotal)
VALUES
('$_POST[custId]','$_POST[itemName]','$_POST[itemPrice]','$_POST[itemQty]','$_POST[itemTotal]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }

if ($sql) { echo "1"; }
else {echo "2"; }

mysql_close($con);
?>

Answer 1

mysql_query("UPDATE table SET sessid = ".session_id()." WHERE user = ".$user_id);

Something like that?