Angular-茉莉花单元测试的模拟Promise方法

Question

测试方法

  public onSubmit(registerData: RegisterDataModel): void {
    this.registrationService.registerWithEmailAndPassword(registerData).then((msg: string[]) =>
      this.router.navigate(['/completeSignUp']).then(() => {
        msg.forEach(singleMessage => this.notificationService.primary(singleMessage));
      }))
      .catch((msg) => msg.forEach(singleMessage => {
        this.notificationService.danger(singleMessage);
      }));
  }

我想测试是否在我的方法中调用了router.navigate。现在，我想嘲笑我的service.registerWithEmailAndPasswort诺言，但以某种方式我无法嘲笑它。

我的规格文件

//Stubs
const routerStub: Router = jasmine.createSpyObj('Router', ['navigate']);
const registryStub: RegistrationService = jasmine.createSpyObj('RegistrationService', ['registerWithEmailAndPassword']);

单元测试

  it('should navigate on promise - success', () => {
    (<jasmine.Spy>registryStub.registerWithEmailAndPassword).and.callThrough();
    const spy = (<jasmine.Spy>routerStub.navigate);
    component.onSubmit({username: null, email: null, password: null, passwordConfirm: null, termsAndCondition: null});
    expect(spy).toHaveBeenCalledWith(['/completeSignUp']);
  });

出现的错误是：TypeError: Cannot read property 'then' of undefined 有人会嘲笑这项服务吗？

编辑

我也曾试图嘲笑诺言，

    (<jasmine.Spy>registryStub.registerWithEmailAndPassword)
  .and.returnValue(new Promise(() => Promise.resolve()));

但是它仍然让我失望：

Expected spy Router.navigate to have been called with [ [ '/completeSignUp' ] ] but it was never called.

Answer 1

正如Silicon Soul所述，您必须使用返回值来模拟router.navigate承诺，否则它将进入Promise.reject()。通过添加(<jasmine.Spy>routerStub.navigate).and.returnValue(Promise.resolve());单元测试应该可以。 最终的单元测试应如下所示：

  it('should navigate on promise - success', fakeAsync(() => {
    const spy = (<jasmine.Spy>routerStub.navigate).and.returnValue(Promise.resolve());
    (<jasmine.Spy>registryStub.registerWithEmailAndPassword).and.returnValue(Promise.resolve(['test']));
    component.onSubmit({username: 'test', email: 'test', password: 'test', passwordConfirm: 'test', termsAndCondition: true});

    tick();
    expect(spy).toHaveBeenCalledWith(['/completeSignUp']);
  }));

Answer 2

您将收到错误消息，因为registerWithEmailAndPassword间谍未返回Promise。您可以使用callFake返回Promise：

(<jasmine.Spy>registryStub.registerWithEmailAndPassword).and.callFake(() => Promise.resolve([]));

此外，promise是异步的，因此您可能应该使用fakeAsync测试和滴答，超时或使用 then 方法而不是Promise返回对象。