无法使用PHP反序列化jQuery数据

Question

I am trying to implement a modal AJAX form. But when I press "Submit" I only receive empty body emails. I guess I have bugs in PHP code.

HTML:

<form class="form" method="POST">
    <input type="text" name="name" placeholder="Name">
    <input type="text" name="phone" placeholder="Phone">
    <input type="submit" value="Submit" >
</form>

Jquery:

$(".form").submit(function() {
    var this1= $(this);
    var form_data = $(this).serialize();
    if (is_empty(this1)){
        $.ajax({
            type: "POST", 
            url: "feedback.php",
            data: form_data,
            success: function (res) {
                swal("Thank you", "very much", "success");
                $('.black_layout').fadeOut(200);
                $('.main_form_wrapper').fadeOut(200);
                this1.trigger('reset');
            }
        });
    }
    return false;
});

PHP in feedback.php file:

parse_str($_POST['form_data'], $formdata);

$name = $formdata['name'];
$phone=$formdata['phone'];
$formcontent="From: $name 
 Phone: $phone";
$recipient = "[e-mail]";
$subject = "MAIL HEADER";
mail($recipient, $subject, $formcontent) or die("Error");

EDIT:I am also not getting a "Thank you" message for some reason.

Answer 1

You have an error during sending form data with ajax request. here is your code which is working fine :

JQuery :

$(".form").submit(function() { 
var this1= $(this);
var form_data = $(this).serialize();
alert(form_data);
//if (is_empty(this1)){
    $.ajax({
        type: "POST", 
        url: "feedback.php",
        data: {'form_data' : form_data },
        success: function (res) { alert(res);
            //swal("Thank you", "very much", "success");
            $('.black_layout').fadeOut(200);
            $('.main_form_wrapper').fadeOut(200);
            this1.trigger('reset');
        }
    });
//}
return false;

});