Exploring Angular unit testing using Jasmine: Techniques to customize or delete spyOn

Current software versions for AngularJS and Jasmine:

AngularJS v1.2.26

Jasmine v2.2.0

I am facing an issue with a spyOn. When attempting to change or remove its behavior, I encounter the following error message: 

Error: getUpdate has already been spied upon

var data1 = 'foo';
var data2 = 'bar';

describe("a spec with a spy", function(){

    beforeEach(module('app'));

    var $q;

    beforeEach(inject(function(_updateService_, _$q_){
        updateService = _updateService_;

        //Spy on the results of the getUpdate()
        $q = _$q_;
        var deferred = $q.defer();
        deferred.resolve( data1 );
        spyOn(updateService, 'getUpdate').and.returnValue(deferred.promise);

    }));

    describe('attempting to use a different spy here', function() {

        it('should return a different value', function() {

          var deferred = $q.defer();
          deferred.resolve( data2 );
          spyOn(updateService, 'getUpdate'); //ERROR OCCURS HERE
          updateService.getUpdate.and.returnValue(deferred.promise);

          ...

        });
    });

...

If I remove the second spyOn statement, the test does not function properly.

What is the correct approach to resolve this issue?

javascriptangularjsunit-testingjasminekarma-jasmine

Answer №1

You have the ability to simply replace it

updateService.updateData = jasmine.createSpy().and.returnValue(etc)

Answer №2

You have the ability to change the spy's result

    var deferred = $q.defer();
    deferred.resolve( newData );

    var fetchSpy = spyOn(fetchService, 'fetchData').and.returnValue(deferred.promise);



    var newDeferred = $q.defer();
    newDeferred.resolve( updatedData );

    fetchSpy.and.returnValue(newDeferred.promise);

Answer №3

Starting from jasmine version 2.5, you can utilize the global allowRespy() configuration.

jasmine.getEnv().allowRespy(true);

If you need to use spyOn() multiple times without accessing the first spy or when it's not required, this setting will allow that. Just keep in mind that it will give back the previous spy if one is already active.

spyOn(updateService, 'getUpdate').and.returnValue(deferred.promise);
...
spyOn(updateService, 'getUpdate').and.returnValue(deferred.promise);

Answer №4

A more convenient and straightforward method:

updateService.getUpdate.and.returnValue(Observable.of({success:true}));

This will give you the result.

Answer №5

Here is an alternative choice:

(testService.operation as jasmine.Spy).and.returnResponse(output);

Answer №6

The green ticked solution didn't do the trick for me, but I found success with this:

yourAwesomeService.addThing = jasmine.createSpy('fake', function(){}).and.returnValue();

While your jasmine test may pass, be prepared for Typescript errors when running your application if you don't include a random string and an empty function as arguments in createSpy().

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