Unique function for ajax request not working

Question

I'd like to create unique function for ajax post request.

This is actual source code:

function doAjax(url, postData) {

  var xhttp = new XMLHttpRequest();
  xhttp.onreadystatechange = function() {
    if (xhttp.readyState == 4 && xhttp.status == 200) {
        return xhttp.responseText;
    }
  };
  xhttp.open("POST", url, true);
  xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
  xhttp.send(postData);

}

console.log( doAjax('shipment.php', 'fname=test1&lname=test2') );

But it's not working, the console shows "undefined".

But, if I change return on alert() - everything is OK. So, why return now working ?

That is how ASYNC works. Do the console.log before the line `return xhttp.responseText;` — mplungjan, Sep 02 '16 at 12:03

score 0 · Answer 1 · answered Sep 02 '16 at 12:04

0

Your function doAjax does not actually return something. When you do xhttp.onreadystatechange = function() {...}; you just set a callback for an event that means this function isn't triggered immediately. No matter what you want to do, you will have to do (or trigger) it inside your callback.

answered Sep 02 '16 at 12:04

Sebastian Kaupper

153
8

It's a FAQ and already closed. – mplungjan Sep 02 '16 at 12:05

Unique function for ajax request not working

1 Answers1