reactjs - 我的模拟函数在 Sagas 测试中没有被调用

Question

我正在尝试用 Jest 测试 Sagas，但我无法让 Sagas 调用 loginApi 模拟函数，它似乎正在调用实际函数。有人可以帮我解决这个问题吗？

这就是我的传奇:

export async function loginApi(user) {
  return await api.post('/auth/customers/login', {
    email: user.email, //<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="98fbedebecf7f5fdeaa9d8f5f1b6f5fd" rel="noreferrer noopener nofollow">[email protected]</a>
    password: user.password, //12345678
  });
}

// Saga function that handles the side effect when the loginWatcher is triggered
export function* loginActionEffect(loginAction) {
  //console.tron.log('login Action Effect');

  try {
    const response = yield call(loginApi, loginAction.user);
    // console.log('response:' + response);

    yield AsyncStorage.setItem('access_token', response.data.access_token);
    yield put(LoginActions.setLogin(loginAction.user));
    yield put(LoginActions.setError(''));
  } catch (e) {
    //console.log('problema com o login!');
    yield put(LoginActions.setError('Email ou senha incorretos!'));
  }
}

这是我的测试:

it('should receive token in case of success', async () => {
    const mockToken = {token: 'kaajfalkfjlaniurvayeg'};
    const loginApi = jest.fn().mockImplementation(() => Promise.resolve(mockToken));
    const dispatched = [];

    const result = await runSaga(
      {
        dispatch: action => dispatched.push(action),
      },
      loginActionEffect,
    );

    expect(loginApi).toHaveBeenCalled();
    expect(dispatched).toContainEqual(Creators.setLogin());
    expect(dispatched).toContainEqual(Creators.setError());
    loginApi.mockClear();
  });

这是我的测试结果:

expect(jest.fn()).toHaveBeenCalled()

Expected number of calls: >= 1
Received number of calls:    0

  21 |     );
  22 | 
> 23 |     expect(loginApi).toHaveBeenCalled();
     |                      ^

Answer 1

runSaga 返回一个 Task 对象，而不是 Promise，因此您需要调用 toPromise

const result = await runSaga(
  {
    dispatch: action => dispatched.push(action),
  },
  loginActionEffect,
).toPromise();

PS。您测试传奇的方式与推荐的方式不符。请阅读此docs 。正如我在评论中所写的。声明性效果的一件很酷的事情是效果执行依赖于解释器。因此，您不需要在测试中实际执行或模拟任何内容。