Angular - Jasmine 單元測試的模擬 Promise 方法

Question

測試方法

  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);
      }));
  }

我想測試是否在我的方法中調用了router.navigate 。 現在我想模擬我的service.registerWithEmailAndPasswort Promise 但不知何故我無法模擬它。

我的規格文件

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

單元測試

  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']);
  });

出現的錯誤是： TypeError: Cannot read property 'then' of undefined人如何正確模擬此服務？

編輯

我也試圖嘲笑這樣的承諾：

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

但它仍然讓我覺得：

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

Answer 1

正如硅靈魂所提到的，您肯定需要使用返回值模擬router.navigate承諾，否則它將進入Promise.reject() 。 通過添加(<jasmine.Spy>routerStub.navigate).and.returnValue(Promise.resolve()); 單元測試應該沒問題。 最終的單元測試應如下所示：

  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

您將收到錯誤消息，因為 registerWithEmailAndPassword 間諜不返回 Promise。 您可以使用 callFake 返回一個 Promise：

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

此外，promise 是異步的，因此您可能應該使用fakeAsync測試和滴答，或超時或使用then方法而不是 Promise 返回對象。