Javascript 使用 Jasmine 进行 Angular 单元测试：如何删除或修改 spyOn
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Angular unit testing with Jasmine: how to remove or modify spyOn
提问by emersonthis
AngularJS v1.2.26
AngularJS v1.2.26
Jasmine v2.2.0
茉莉花 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
如何更改或删除 a 的行为spyOn？当我尝试覆盖它时，出现以下错误：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.
当我删除第二个 spyOn 时，测试不起作用。
How do I do this?
我该怎么做呢？
回答by Peter Ashwell
You can just overwrite it
你可以覆盖它
updateService.getUpdate = jasmine.createSpy().and.returnValue(etc)
回答by Javier Candalaft-CONT
You can override the return value of the spy
您可以覆盖间谍的返回值
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);
回答by André Werlang
Since jasmine v2.5, use the global allowRespy()setting.
从 jasmine v2.5 开始，使用全局allowRespy()设置。
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()当您不想和/或有权访问第一个间谍时，您将能够多次呼叫。当心它会返回之前的间谍，如果任何已经处于活动状态。
spyOn(updateService, 'getUpdate').and.returnValue(deferred.promise);
...
spyOn(updateService, 'getUpdate').and.returnValue(deferred.promise);回答by Aniruddha Das
More easier way is to simple
更简单的方法是简单
updateService.getUpdate.and.returnValue(Observable.of({status:true}));
回答by rb1econ
the green check-marked answer didn't work for me, but this did:
绿色复选标记的答案对我不起作用，但这确实：
yourCoolService.createThing = jasmine.createSpy('notreal', function(){}).and.returnValue();
your jasmine test will run but when you go to fire up your app typescript will yell loudly at you if you don't put a random string and an empty function as the args to createSpy().
您的 jasmine 测试将运行，但是当您启动时，如果您不将随机字符串和空函数作为 args 放置到createSpy().
回答by Justas
Another option:
另外一个选项：
(yourService.method as jasmine.Spy).and.returnValue(value);
