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?
You can just overwrite it
updateService.getUpdate = jasmine.createSpy().and.returnValue(etc)
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);
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);