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如何以正确的顺序从Redux中间件调度操作

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作为世界上应用程序的必备品,我的React应用程序需要对API执行一些Ajax调用 .

我选择使用Redux中间件来正确地将API提取与我的组件分开 .
我的想法是从我的组件发送 REQUEST 动作 . 中间件监听它们并发送 SUCCESSERROR 动作:这些动作最后由reducers监听 .

这里有很多人已经问过如何从Redux中间件调度操作:这是我在这里问题的主题:)

首先,让我向您展示一个我用来编写的简单减速器:

function currentUserReduxer(state = {}, action = {}) {
  const { currentUser, error } = action.payload;

  switch (action.type) {
    case 'GET_CURRENT_USER_REQUEST':
      return { ...state, isFetching: true, error: null };

    case 'GET_CURRENT_USER_SUCCESS':
      return { ...state, id: currentUser.id, isFetching: false };

    case 'GET_CURRENT_USER_FAILURE':
      return { ...state, isFetching: false, error };

    default:
      return state;
  }
}

和相应的中间件:

() => next => async (action) => {
  next(action);

  switch (action.type) {
    case'GET_CURRENT_USER_REQUEST': {
      try {
        const currentUser = await api.getCurrentUser();
        next(actions.getCurrentUserSuccess(currentUser));
      } catch (error) {
        next(actions.getCurrentUserFailure(error));
      }
      break;
    }

    default:
      break;
  }
};

它很长时间以来运行良好,然后我意识到它部分错误:我的中间件没有返回 next 的值,所以它打破了中间件链,这是错误的!
由于 next(action); 是我've performed in the middleware, I couldn'牛逼很快回到它本的第一件事,所以我've moved it at the end of the middleware. I'以前也决定派遣新的行动,而不是使用 next 他们(毕竟,他们是新的行动,这有一定道理送他们到整个产业链中间件) . 我的新中间件现在看起来像这样:

store => next => async (action) => {
  switch (action.type) {
    case'GET_CURRENT_USER_REQUEST': {
      try {
        const currentUser = await api.getCurrentUser();
        store.dispatch(actions.getCurrentUserSuccess(currentUser));
      } catch (error) {
        store.dispatch(actions.getCurrentUserFailure(error));
      }
      break;
    }

    default:
      break;
  }

  return next(action);
};

它看起来很棒,但我现在有另一个问题:因为 store.dispatch 是同步的,所以 next(action) 被调用 . That means that my reducers receive the REQUEST action AFTER the SUCCESS or FAILURE ones :(

我认为一个解决方案可能是使用旧的承诺而不是 await

store => next => async (action) => {
  switch (action.type) {
    case'GET_CURRENT_USER_REQUEST': {
      api.getCurrentUser()
        .then((currentUser) => {
          store.dispatch(actions.getCurrentUserSuccess(currentUser));
        })
        .catch((error) => {
          store.dispatch(actions.getCurrentUserFailure(error));
        });
      break;
    }

    default:
      break;
  }

  return next(action);
};

另一个想法是用 setTimeout 包装 store.dispatch

store => next => async (action) => {
  switch (action.type) {
    case'GET_CURRENT_USER_REQUEST': {
      try {
        const currentUser = await api.getCurrentUser();

        setTimeout(() => store.dispatch(actions.getCurrentUserSuccess(currentUser)));
      } catch (error) {
        setTimeout(() => store.dispatch(actions.getCurrentUserFailure(error)));
      }
      break;
    }

    default:
      break;
  }

  return next(action);
};

说实话,我真的不喜欢这两种解决方案,他们觉得这么hacky ......

所以这是我的问题:我怎么想处理我的问题?有更干净的方法吗?

提前致谢 :)

1 回答

  • 3

    好像你想要做的事情类似于Redux-Saga我建议你去看看他们的图书馆 .

    取自他们的榜样

    // worker Saga: will be fired on USER_FETCH_REQUESTED actions
    function* fetchUser(action) {
       try {
          const user = yield call(Api.fetchUser, action.payload.userId);
          yield put({type: "USER_FETCH_SUCCEEDED", user: user});
       } catch (e) {
          yield put({type: "USER_FETCH_FAILED", message: e.message});
       }
    }
    
    /*
      Starts fetchUser on each dispatched `USER_FETCH_REQUESTED` action.
      Allows concurrent fetches of user.
    */
    function* mySaga() {
      yield takeEvery("USER_FETCH_REQUESTED", fetchUser);
    }
    

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