用AJAX调用的PHP函数不起作用

Question

I am attempting to save favorites to a session by clicking on a button. I'm using AJAX to call the php function save_favorite(). The function is being called but for some reason the save_favorite_in_session_variable() method just won't work when being called like this. I've tested the function by itself and it works fine. Does anyone have a clue what I'm missing?

//JavaScript

$(document).on('click', '.add-favorite', function() {
    var imgId = $(this).data('id');
    saveFavorite(imgId);
});

function saveFavorite(imgId) {
    var data = {
        'action': 'save_favorite',
        'id' : imgId
    };
    jQuery.post(my_ajax_object.ajax_url, data, function(response) {
        alert(response);
    });
}

// PHP

function save_favorite() {
    ob_start();
    debug_to_console('saving favorite');

    if ($_POST['id']) {
        $post_id = $_POST['id'];
        save_favorite_in_session_variable($post_id);
    }
    wp_die();
}
add_action('wp_ajax_save_favorite', 'save_favorite');
add_action('wp_ajax_nopriv_save_favorite', 'save_favorite');

function save_favorite_in_session_variable($post_id) {
    array_push($_SESSION['favorites'], $post_id);
}

Answer 1

Maybe too simple, but much forgotten; dit you use 'session_start()' in the PHP file? (To initiate the session manager)

Also, always first check if $_SESSION['favorites'] is available and an array with:

if(isset($_SESSION['favorites']) && is_array($_SESSION['favorites'])){
    array_push($_SESSION['favorites'], $post_id);
}

To avoid errors, it's easier to handle it directly instead of after the fix.