Getting product_id - Woocommerce - Ajax pop-up

Question

I have this code below, but just don't seem to managed to get "product_id" into html to show the right product. Can somebody point me into the right direction?

I'm new to Ajax, so just don't know how to fetch data from ajax into html.

How could I get "product_id" into this form: < div > echo $product_id < /div > ?

<script>
jQuery(document).ready(function($) {
var modal = document.getElementById('myModal');
var span = document.getElementsByClassName("close")[0];

span.onclick = function() {
    modal.style.display = "none";
}
window.onclick = function(event) {
    if (event.target == modal) {
        modal.style.display = "none";
    }
}
// Let's do the magic
    // make sure to change #myBtn to read more button selector
    $('.products > li .add_to_cart_button').click(function(e) {
    e.stopImmediatePropagation();
    e.preventDefault();
    // get url
        var url = $(this).attr('href');
        var container = $('#myModal').find('.popup');
        var data = {
            action: 'show_product',
            url: url,

        };
        $.post('<?php echo esc_url( home_url() ); ?>/wp-admin/admin-ajax.php', data, function(response) {
           // display the response
           console.log(response);
           $(container).empty();       
           $(container).html(response);
            modal.style.display = "block";
        });
    });
});
</script>
<?php
}
add_action( 'wp_footer', 'pop_up' );


add_action('wp_ajax_show_product', 'show_product_callback_wp');
add_action( 'wp_ajax_nopriv_show_product', 'show_product_callback_wp' );
function show_product_callback_wp() {
     $url = $_POST['url'];
     $product_id = url_to_postid( $url );

    // product content
    $content_post = get_post($product_id);
    $content = $content_post->post_content;
    $content = apply_filters('the_content', $content);
    $content = str_replace(']]>', ']]&gt;', $content);

     $output = "";
     $output .= get_the_post_thumbnail( $product_id, 'medium');;
     $output .= $content;

     echo $output;
     exit(); 
}

Answer 1

等待大神答复