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

Answer 1

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/