Question

我创建了一个购物车脚本，以便使用PHP和MySQL添加或删除产品。该脚本按预期工作，但它不是很舒服。我试图将它与ajax调用一起使用。

1。我在购物车中添加了所选产品

$('#add-product').on('click',function() {  
$.ajax({
  var product_id = $(this).data('id'); //Product id
  url: 'cart.php',
  data:{add:product_id },
  type:'GET',
  success: function(data) {

     if(!data.error){
      $('#show-cart').html(data);
     }
       }

          });
             });

2。 cart.php

         <?php
        require_once('resource/init.php');


        //ADD A PRODUCT IN THE SHOPPING CART
        if(isset($_GET['add'])) {

        $query = query("SELECT * FROM products WHERE id = ".escape_string($_GET['add']));
        confirm($query);

        while ($row = fetch_array($query)) {
        if($row['quantity'] !=  $_SESSION['product_'.$_GET['add']]){
        $_SESSION['product_'.$_GET['add']] +=1;
        }
        }
        }


        //DELETE PRODUCT FROM SHOPPING CART
        if(isset($_GET['delete'])) {
        $_SESSION['product_'.$_GET['delete']] = '0';
        unset($_SESSION['total']);
        unset($_SESSION['quantity']);
        }



        $total = 0;
        $item_quantity = 0;
        $item_name = 1;
        $item_number = 1;
        $amount = 1;
        $quantity = 1;


        foreach ($_SESSION as $name => $value) {


        if(substr($name, 0,8) == 'product_' && $value > 0){
        $lenght =strlen($name);
        $id = substr($name, 8, $lenght);
        $query = query("SELECT * FROM products WHERE id = ".escape_string($id)." ");
        confirm($query);

        while($row = fetch_array($query)) {

        $sub_price = pr_get_sub_price($row['price'],$row['discount'],$row['id'],$value,$lang);
        $item_quantity += $value;
        $product_image = 'images/products/'.$row['image'];
        $price = show_product_price($row['price'],$row['discount'],$lang);
        $discounted_price = get_product_price_by_discount($row['price'],$row['discount'],$row['id'],$lang);



        ?>

        <!--THIS IS THE SHOPPING CART-->
        <table>
        <tr style="text-align:left">
        <td><a href="#"><img src="<?php echo $product_image;?>" alt="cart-hover" style="float:left;width: 58px; margin-right:14px;"/></a></td>
        <td><a title="<?php echo $row['title'];?>" href="#" style="margin-right:10px;"><span class="cart-title"><?php echo $row['title'];?> (<?php echo $value;?>)</span></a></td>
        <td><span style="font-size: 90%; margin-right: 8px; position:relative;top:2px;"><?php echo $price;?> <?php echo $discounted_price;?></span></td>

        <!--DELETE BUTTON-->
        <td>
        <button data-id = "<?php echo $row['id'];?>" class="delete-product">   <i class="fa fa-close"></i> Delete</button></td>
        </tr>
        </table>





        <input type="hidden" name="item_name_<?php echo $item_name;?>" value="<?php echo $row['title'];?>">
        <input type="hidden" name="item_number_<?php echo $item_number;?>" value="<?php echo $row['id'];?>">
        <input type="hidden" name="amount_<?php echo $amount;?>" value="<?php echo $row['price'];?>">
        <input type="hidden" name="quantity_<?php echo $quantity;?>" value="<?php echo $value;?>">

        <?php
        $item_name++;
        $item_number++;
        $amount++;
        $quantity++;

        $_SESSION['total'] = $total += $sub_price;
        $_SESSION['quantity'] = $item_quantity;

        }
        }
        }



        ?>


   <!--DELETE PRODUCT-->
    <script>

    $('.delete-product').click(function() {

    var product_id = $(this).data('id');//PRODUCT ID

    $.ajax({
         url: 'cart.php?delete=' + product_id,
         type: 'get',
         success:function(data){
         $('#show-cart').html(data); 

        }

         });

           });
     </script>

上述脚本按预期工作，但不适用于DELETE操作。

如果我有购物车中的产品： 产品一（3）$ 250.00

然后点击DELETE图标，所选产品继续显示在购物车中，因为它在开始之前点击DELETE图标：产品一（3）$ 250.00

如果我尝试在购物车中添加相同的产品，则会以值1开头显示：产品一（1）$ 250.00

这是我第一次使用PHP / MySQL和Ajax，我不知道如何解决这个问题。

如何使用PHP / MySQL和Ajax在购物车中添加/删除产品

0 个答案: