Understanding Problem - Jquery Deferred chaining

Question

function ajax_call() {

    var ajaxCallMock = $.Deferred().resolve('A');

    return ajaxCallMock.done(function(response) {

            return 'B';

    });
}



ajax_call().done(function(response) {
    console.log(response);
});

I would expect the console output to be 'B', but I get 'A'. Why?

I think a resolved promise always returns another solved promise, is that wrong? — Lokomotywa
– Lokomotywa, Commented Apr 9, 2019 at 12:53
jQuery's .done() is not a synonym for .then(). There are few if any cases where you would use .done(). — Roamer-1888
– Roamer-1888, Commented Apr 9, 2019 at 18:02

charlietfl · Accepted Answer · 2019-04-09 12:59:06Z

2

Use then() instead of done() as done doesn't return a new promise.

function ajax_call() {

  var ajaxCallMock = $.Deferred().resolve('A');

  return ajaxCallMock.then(function(response) {
    return 'B';
  });
}



ajax_call().then(function(response) {
  console.log(response);
});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

edited Apr 9, 2019 at 12:59

answered Apr 9, 2019 at 12:53

charlietfl

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1 Comment

Lokomotywa Over a year ago

Can I simply replace the var ajaxCallMock = $.Deferred().resolve('A'); with a $.get() Ajax Call? and still use then?

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