首先，在应用程序启动时，减速机的状态是新鲜的，并带有默认的InitialState。

我们必须添加一个动作，调用APP初始加载持续默认状态。

当注销应用程序时，我们可以简单地重新分配默认状态，reducer将像新的一样工作。

APP主容器

  componentDidMount() {   
    this.props.persistReducerState();
  }

主APP减速器

const appReducer = combineReducers({
  user: userStatusReducer,     
  analysis: analysisReducer,
  incentives: incentivesReducer
});

let defaultState = null;
export default (state, action) => {
  switch (action.type) {
    case appActions.ON_APP_LOAD:
      defaultState = defaultState || state;
      break;
    case userLoginActions.USER_LOGOUT:
      state = defaultState;
      return state;
    default:
      break;
  }
  return appReducer(state, action);
};

注销时调用重置状态的动作

function* logoutUser(action) {
  try {
    const response = yield call(UserLoginService.logout);
    yield put(LoginActions.logoutSuccess());
  } catch (error) {
    toast.error(error.message, {
      position: toast.POSITION.TOP_RIGHT
    });
  }
}

只是对@dan-abramov答案的扩展，有时我们可能需要保留某些被重置的键。

const retainKeys = ['appConfig'];

const rootReducer = (state, action) => {
  if (action.type === 'LOGOUT_USER_SUCCESS' && state) {
    state = !isEmpty(retainKeys) ? pick(state, retainKeys) : undefined;
  }

  return appReducer(state, action);
};

使用Redux Toolkit的方法:

export const createRootReducer = (history: History) => {
  const rootReducerFn = combineReducers({
    auth: authReducer,
    users: usersReducer,
    ...allOtherReducers,
    router: connectRouter(history),
  });

  return (state: Parameters<typeof rootReducerFn>[0], action: Parameters<typeof rootReducerFn>[1]) =>
    rootReducerFn(action.type === appActions.reset.type ? undefined : state, action);
};

我发现Dan Abramov的回答很适合我，但它触发了ESLint no-param-reassign错误- https://eslint.org/docs/rules/no-param-reassign

下面是我如何处理它，确保创建一个状态的副本(这是，在我的理解，Reduxy的事情要做…):

import { combineReducers } from "redux"
import { routerReducer } from "react-router-redux"
import ws from "reducers/ws"
import session from "reducers/session"
import app from "reducers/app"

const appReducer = combineReducers({
    "routing": routerReducer,
    ws,
    session,
    app
})

export default (state, action) => {
    const stateCopy = action.type === "LOGOUT" ? undefined : { ...state }
    return appReducer(stateCopy, action)
}

但是也许创建一个状态的副本，然后把它传递给另一个减速器函数，它会创建一个状态的副本，这有点过于复杂了?这篇文章读起来不太好，但更切题:

export default (state, action) => {
    return appReducer(action.type === "LOGOUT" ? undefined : state, action)
}

结合Dan Abramov的回答，Ryan Irilli的回答和Rob Moorman的回答，来解释保持路由器状态和初始化状态树中的其他所有东西，我最终得到了这样的答案:

const rootReducer = (state, action) => appReducer(action.type === LOGOUT ? {
    ...appReducer({}, {}),
    router: state && state.router || {}
  } : state, action);

对我来说，最好的工作是设置initialState而不是state:

  const reducer = createReducer(initialState,
  on(proofActions.cleanAdditionalInsuredState, (state, action) => ({
    ...initialState
  })),