ajax提交停留在同一页上+更新购物车中产品的数量

Question

Hello I use ajax to stay on the same page after sumbit form. and if success I will use ajax load to load other page that has a text about quantity of product. the problem is the if I first submit, the quantity +1/ second submit the quantity +2 (so total 3 now) / third submit, the quantity +3 (so total 6 now). I want it to +1 only

I have 3 page.

product.php page (have a products and button add-to-cart(submit) ) I use ajax here

$(document).ready(function() {
    $('.add-to-cart').on('click', function() {

        $(document).on('submit', '.myForm', function(e) {
            $.ajax({
                url: $(this).attr('action'),
                type: $(this).attr('method'),
                data: $(this).serialize(),
                success: function(html) {
                    $("#outside_cart_text").load("<?php echo base_url();?>shopping/ajax_text.php", function(responseTxt, statusTxt, xhr) {
                        if (statusTxt == "success")

                            if (statusTxt == "error")
                            alert("Error: " + xhr.status + ": " + xhr.statusText);
                    });
                }
            });
            e.preventDefault();
        });

    });
});

A page that ajax load if success shopping/ajax_text.php

<?php echo $_SESSION('qty_type'); ?> type <?php echo $_SESSION('qty_product');?> 

The cart page - the first page(product.php) form will submit to this page. and this page has code logic about add item to cart. so I create session in this page

$_SESSION['qty_type'] = $sizeof_cart_array;
$_SESSION['qty_product'] = $total_item;

Answer 1

why you want to call on click and on submit as well at the same time ? Just invoke only on click and handle it.

$(document).ready(function() { 

           $('.add-to-cart').on('click', function() {


           $(document).on('submit', '.myForm', function(e) {<br/>

              /// code goes here