Angular unit testing with Jasmine: how to remove or modify spyOn

Question

AngularJS v1.2.26

Jasmine v2.2.0

How can I change or remove the behavior of a spyOn? When I try to override it, I get the following error: 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 the results of the getUpdate()
        $q = _$q_;
        var deferred = $q.defer();
        deferred.resolve( data1 );
        spyOn(updateService, 'getUpdate').and.returnValue(deferred.promise);

    }));

    describe('and here the spy should be different', function() {

        it('returns a different value', function() {

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

          ...

        });
    });

...

When I remove the second spyOn the test doesn't work.

How do I do this?

Solution 1

You can just overwrite it

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

Solution 2

You can override the return value of the spy

    var deferred = $q.defer();
    deferred.resolve( data1 );

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



    var newDeferred = $q.defer();
    newDeferred.resolve( data2 );

    getUpdateSpy.and.returnValue(newDeferred.promise);        

Solution 3

Since jasmine v2.5, use the global allowRespy() setting.

jasmine.getEnv().allowRespy(true);

You'll be able to call spyOn() multiple times, when you don't want and/or have access to the first spy. Beware it will return the previous spy, if any is already active.

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

Solution 4

More easier and simpler way:

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

It will return the value.

Solution 5

Another option:

(yourService.method as jasmine.Spy).and.returnValue(value);

View more solutions

Comments

[Bohdan Lyzanets]

Possible duplicate of Jasmine SpyOn same method more than once

[emersonthis]

Is there way to remove the spy entirely? To go back to the original function?

[Jan]

The big question is, if you don't the same functionality in every test, why have a global spy at all? If you want to set the spy for every test, then set the spy for every test.

[theblang]

@Jan If I have fifty tests, and only one of them has a spy for a function that is different from the rest, I'd rather just change it the one time there instead of every single test.

[FlavorScape]

well, in my case it's on a global object (defined by native code). So I have multiple specs that need to return different values in different situations....

[MBielski]

This is probably due to an updated version of Jasmine. I am using 2.7 and this works for me.

[camleng]

Thank you. Everyone else is missing the obvious answer.

[George C.]

Works on 18/6/2019

[Novastorm]

Honestly, this is the answer I've been looking for ages. I wish it was more prevalent on the internet. A million thanks :)

[GarfieldKlon]

That may work if all your code is in the same describe, but otherwise not because getUpdateSpy is not defined...

[aj go]

would it return to its original implementation by calling jasmine.createSpy().and.callThrought()?

[Samuel Bushi]

@GarfieldKlon You can work around that by defining the variable getUpdateSpy in the parent describe block.

[ANeves]

It seems that your type system is not complaining. That begs the question: why use Typescript at all?