How to fake jquery.ajax() response?

Question

I am writing some QUnit tests for a JavaScript that makes AJAX calls.

For isolation I overwrite $.ajax to write the parameter array of an AJAX call to a variable. This works to test how methods use AJAX functions, but I have difficulty testing the success handler of $.load()

From the documentation at http://api.jquery.com/load/:

When a successful response is detected (i.e. when textStatus is "success" or "notmodified"), .load() sets the HTML contents of the matched element to the returned data.

So I have attempted to return an object containing objects with the same name as variables for the success handler:

    //Mock ajax function
    $.ajax = function (param) {
        _mockAjaxOptions = param;
        var fakeAjaxSuccess = { responseText: "success", textStatus: "success", XMLHttpRequest: "success" };
        return fakeAjaxSuccess;
    };

But this approach hasn't worked.

How can I replicate the behaviour of a successful AJAX call?

Answer 1

This question has a few years and for the new versions of jQuery has changed a bit.

To do this with Jasmin you can try Michael Falaga's approach

Solution

  function ajax_response(response) {
    var deferred = $.Deferred().resolve(response);
    return deferred.promise;
  }

With Jasmine

  describe("Test test", function() {
    beforeEach(function() {
      spyOn($, 'ajax').and.returnValue(
        ajax_response([1, 2, 3])
      );
    });
    it("is it [1, 2, 3]", function() {
      var response;
      $.ajax('GET', 'some/url/i/fancy').done(function(data) {
        response = data;
      });
      expect(response).toEqual([1, 2, 3]);
    });
  });

No Jasmine

  $.ajax = ajax_response([1, 2, 3]);
  $.ajax('GET', 'some/url/i/fancy').done(function(data) {
     console.log(data); // [1, 2, 3]
  });
 

Answer 2

After reading inspired by @Robusto and @Val, I found a method that works:

//Mock ajax function
$.ajax = function (param) {
    _mockAjaxOptions = param;
    //call success handler
    param.complete("data", "textStatus", "jqXHR");
};

Instead of raising the event from any real $.ajax code or by triggering any events, I have my fake ajax object call the function (which is passed in as a parameter to $.ajax()) as part of my fake function.

Answer 3

Use a closure to override $.ajax with a dummy response

After trying the accepted answer and the answer posted by user1634074, I devised this simple and flexible blend of the two.

In its most basic form…

function ajax_response(response) {
  return function (params) {
    params.success(response);
  };
}
$.ajax = ajax_response('{ "title": "My dummy JSON" }');

In the above example, define a function ajax_response() that accepts some JSON string as an argument (or any number of custom arguments useful for simulating a response) and returns an anonymous closure function that will be assigned to $.ajax as an override for unit testing.

The anonymous function accepts a params argument which will contain the settings object passed to the $.ajax function. And it uses the argument(s) passed to the outer function to simulate a response from the server. In this example, it always simulates a successful response from the server, by simply invoking the success callback and supplying it with the dummy JSON.

It is easy to reconfigure…

function ajax_response(response, success) {
  return function (params) {
    if (success) {
      params.success(response);
    } else {
      params.error(response);
    }
  };
}

// Simulate success
$.ajax = ajax_response('{ "title": "My dummy JSON." }', true); 
doAsyncThing(); // Function that calls $.ajax

// Simulate error
$.ajax = ajax_response('{ "error": "Who is the dummy now?" }', false); 
doAsyncThing(); // Function that calls $.ajax

Below we can see it in action…

/* FUNCTION THAT MAKES AJAX REQUEST */
function doAsyncThing() {
  $.ajax({
    type: "POST",
    url: "somefile.php",
    // data: {…},
    success: function (results) {
      var json = $.parseJSON(results),
          html = $('#ids').html();
      $('#ids').html(html + '<br />' + json.id);
    }
  });
}

/* BEGIN MOCK TEST */
// CREATE CLOSURE TO RETURN DUMMY FUNCTION AND FAKE RESPONSE
function ajax_response(response) {
  return function (params) {
    params.success(response);
  };
}

var n = prompt("Number of AJAX calls to make", 10);

for (var i = 1; i <= n; ++i) {
  
  // OVERRIDE $.ajax WITH DUMMY FUNCTION AND FAKE RESPONSE
  $.ajax = ajax_response('{ "id": ' + i + ' }');
  doAsyncThing();
}
/* END MOCK TEST */

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<p id="ids">IDs:</p>

Answer 4

Mock $.ajax as needed without disturbing jQuery

The answers here are good but had a specific need to build out a fake response to a single API call while leaving all other API calls the same until the backend service was built out so I can continue building stuff on the UI.

The API object uses $.ajax under the hood so you can call an API method like so:

api.products({ price: { $lt: 150, tags: ['nike', 'shoes'] } })
.done(function(json) {
  // do something with the data
})
.error(function(err) {
  // handle error
});

This method does the trick:

function mockAjax(options) {
  var that = {
    done: function done(callback) {
      if (options.success)
        setTimeout(callback, options.timeout, options.response);
      return that;
    },
    error: function error(callback) {
      if (!options.success)
        setTimeout(callback, options.timeout, options.response);
      return that;
    }
  };
  return that;
}

Then override a single api call without touching $.ajax:

api.products = function() {
  return mockAjax({
    success: true,
    timeout: 500,
    response: {
      results: [
        { upc: '123123', name: 'Jordans' },
        { upc: '4345345', name: 'Wind Walkers' }
      ]
    }
  });
};

https://jsfiddle.net/Lsf3ezaz/2/

Answer 5

Look at the jQuery documentation: You'll see that the Ajax setup provides a number of other conditions that are tested for. If you make them all point to your fakeAjaxSuccess, you might achieve for your objective.

Alternatively, wrap your $.ajax call into its own function and have whatever calls it simply call your event handler with the fakeAjaxSuccess object.

Answer 6

I think the link below should help. as for a parameter I am not so sure but it could be .

$.fn.ajax.success =  function (){
  ///the rest goest here
}

Override jQuery .val() function?

Answer 7

Here is a simple working solution

var set_ajax_response = function(data){
  $.ajax = $.Deferred().resolve(data).promise;
}

var data = [1,2,3];

set_ajax_response(data);