PHP-Jquery-AJAX，显示弹出消息

Question

I lost my path to finding the solution and need help immediately. I have a php page called list.php which shows list of products with BUY icon on each product. Once the BUY button clicked, it will directing to action.php (which contain script to add into the database) and then return into the same page using header(location:list.php?id=$r[id_product]). I want to show a popup message "the product is now added to cart" after returned into the list.php. How to do that?

Answer 1

You can you cookie and after returning just check if there is specified cookie or not and if there is, show message.

However if you using ajax you don't need a cookie, just add pop-up function in ajax parsing function, something like that

function addToBasket(itemId) {
    var x = new XMLHttpRequest();
    x.open('GET', '/ajax/action.php?add=' + itemId, true);
    x.onreadystatechange = function() {
        if (x.readyState == 4) && (x.status == "200") {
           var responseObj = eval('(' + x.responseText + ')');
           if (responseObj.status == 0)
                window.alert('The product successfully added to your basket');
           else 
                window.alert('An error has occured during the processing the data. Please try again later');
        }
    };
    x.send(null); 

action.php code example

<?php
    // ...

    $response = array('status' => 1);

    if (isset($_GET['itemId'] && $_GET['itemId'] != '')
    {
        $item_id = htmlspecialchars($_GET['itemId']);
        if (is_numeric($item_id) 
        {
            put_in_table($item_id);
            $response = array('status' => 0);
        } 
    }

    // output an Ajax response
    header('Content-Type: application/javascript');
    echo json_encode($response);
?>