代码之家 › 专栏 › 技术社区 › MarcoLe

茉莉花单元试验的角模拟承诺法

karma-jasmine mocking jasmine unit-testing angular

8

MarcoLe · 技术社区 · 8 年前

试验方法

  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);
      }));
  }

我想测试一下 router.navigate 在我的方法中被调用。现在我想嘲笑我 service.registerWithEmailAndPasswort 答应我,但我不能嘲笑。

我的规范文件

//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 有谁能恰当地模仿这项服务吗?

编辑

我也曾试图嘲弄这个承诺,比如:

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

但它仍然让我:

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

2 回复 | 直到 8 年前

1

8

user9353955user9353955 8 年前

正如silicon Soul所说,你需要明确地嘲笑 router.navigate 有回报价值的承诺,否则它将成为 Promise.reject() . 通过添加 (<jasmine.Spy>routerStub.navigate).and.returnValue(Promise.resolve()); 单元测试应该没问题。最后的单元测试应该是:

  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']);
  }));

2

1

SiliconSoul 8 年前

您将得到错误,因为registerWithEmailAndPassword间谍没有返回承诺。你可以用callFake来回信:

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

而且,承诺是异步的,所以您可能应该使用 fakeAsync 测试并勾选,或超时或返回对象然后方法而不是承诺。