问AJAX成功函数从未实现

EN

// this is the id of the form
$("#password_form").submit(function(e) {

    e.preventDefault();
    $(".verify-user-loader").addClass("force-display-block");
    password = $("input#password-reset").val();
    url = 'reset-pass.php';

    $.ajax({

        type : 'POST',
        url : url,
        dataType : 'json',
        data : {
                cardnumber: <?php echo '\''.$cardnumber.'\''; ?>,
                act_token: <?php echo '\''.$activationToken.'\''; ?>,
                password: password
        },
        success : function(success){
            alert("success");
        },
        error : function(request, error) {
            console.log(arguments);
            alert(" Can't do because: " + error);
        }

    });

});

include_once '/../login/user.class.php';


$activationToken = $_POST['act_token'];
$cardnumber = $_POST['cardnumber'];
$password = $_POST['password'];

$user = new User();

$verifyToken = $user->verifyToken( $activationToken, $cardnumber );

if ($verifyToken['status'] === true) {

    $tokenStatus = "inactive";
    $user->signUp( $cardnumber, $password );
    $user->changeTokenStatus( $cardnumber, $tokenStatus );

    $success = true;

    return $success;

}else{
    print_r($verifyToken);
}