如何使用redux-persist保持React-native redux状态？

Question

我一直在尝试使用redux-perist将我的redux-state保存到AsyncStorage。虽然我一直收到错误：

_this.store.getState is not a function

我不确定为什么会这样？

这是我的设置：

configureStore.js：

import {AsyncStorage,} from 'react-native';
import { createStore, applyMiddleware, compose, combineReducers, } from 'redux';
import reduxThunkMiddleware from 'redux-thunk';
import Reactotron from 'reactotron';
import * as reducers from './modules';
import devTools from 'remote-redux-devtools';
import {persistStore, autoRehydrate} from 'redux-persist'


Reactotron.connect({
  enabled: __DEV__,
});

const enhancer = compose(
  autoRehydrate(),
  applyMiddleware(
    reduxThunkMiddleware,
    Reactotron.reduxMiddleware,
  ),
  devTools()
);

export default function configureStore(initialState) {
  const store = createStore(
    combineReducers({
      ...reducers,
    }),
    initialState,
    enhancer,
  );
  Reactotron.addReduxStore(store, {storage: AsyncStorage});
  return store;
}

App.js：

以下是我将store与<provider>：

相关联的位置

import React from 'react';
import {AsyncStorage} from 'react-native';
import { Provider, connect } from 'react-redux';
import { Router } from 'react-native-router-flux';
import routes from '@routes/app';
import createStore from './redux/create'; // Returns store object from the above configureStore.js!
import {persistStore} from 'redux-persist'
const RouterWithRedux = connect()(Router);

const store = persistStore(createStore(), {storage: AsyncStorage}); // Not working??
const Kernel = () => (
  <Provider store={store}>
    <RouterWithRedux scenes={routes} />
  </Provider>
);

export default Kernel;

Answer 1

const RouterWithRedux = connect()(Router);
const store = createStore();
const persistor = persistStore(store, {storage: AsyncStorage}); // Not working??
const Kernel = () => (
  <Provider store={store} persistor={persistor}>
    <RouterWithRedux scenes={routes} />
  </Provider>
);

问题是我必须传递一个persistor字段以及store字段。

在加入persistor字段后，我的商店被持久化为AsyncStorage

编辑：

当时有效 - 我发现这不是正确解决问题的方法。但我仍然得到它仍然有效的回应，如果有人可以为其他人提供另一个答案，那就太好了。