How to properly return result in php/ajax?

Question

AJAX

Good afternoon. There is the following code:

function addPosts() {
            $.ajax({
                type: 'post',
                url: "/example.php",
                cache: false,
                response: 'text',
                success: function(data) {
                    $("#getNotice").html(data);
                }
            });
        }

After returning the data in #getNotice , you need to hide the data after a while. Tried delay(3000).fadeOut() but the whole problem is that after hiding they don't show up anymore.
It works as follows: the user clicks on the button and receives a notification that his data was successfully added, but when he clicks again, the notification does not come.
I considered show() and hide() , but it seems to me that I'm looking in the wrong place and not doing it quite right.
Thank you!

Answer 1

it's simple, you didn't think of it

success: function(data) {
                    $("#getNotice").html(data);
                    $("#getNotice").show(); // на уже показанный объект никак не влияет
                    $("#getNotice").fadeOut(3000); // фэйд в течение 3 сек.
                }

Answer 2

You looked right, just hide it by timeout.

Answer 3

setTimeout(function(){
$("#getNotice").addClass('hidden'); // добавить класс, который display: none;
}, 3000);