为了避免Redux引用初始状态的相同变量，我的建议是:

// write the default state as a function
const defaultOptionsState = () => ({
  option1: '',
  option2: 42,
});

const initialState = {
  options: defaultOptionsState() // invoke it in your initial state
};

export default (state = initialState, action) => {

  switch (action.type) {

    case RESET_OPTIONS:
    return {
      ...state,
      options: defaultOptionsState() // invoke the default function to reset this part of the state
    };

    default:
    return state;
  }
};

一种方法是在应用程序中编写一个根减速器。

根减速机通常会将处理操作委托给combineReducers()生成的减速机。但是，无论何时它接收到USER_LOGOUT操作，它都会再次返回初始状态。

例如，如果你的根减速器是这样的:

const rootReducer = combineReducers({
  /* your app’s top-level reducers */
})

你可以将它重命名为appReducer，并编写一个新的rootReducer委托给它:

const appReducer = combineReducers({
  /* your app’s top-level reducers */
})

const rootReducer = (state, action) => {
  return appReducer(state, action)
}

现在我们只需要教新的rootReducer返回初始状态以响应USER_LOGOUT操作。如我们所知，无论操作如何，当调用以undefined作为第一个参数时，约简器都应该返回初始状态。让我们用这个事实来有条件地剥离累积状态，当我们把它传递给appReducer:

 const rootReducer = (state, action) => {
  if (action.type === 'USER_LOGOUT') {
    return appReducer(undefined, action)
  }

  return appReducer(state, action)
}

现在，每当USER_LOGOUT触发时，所有减约器都将重新初始化。它们还可以返回与初始值不同的值，因为它们可以检查动作。也要打字。

重申一下，完整的新代码是这样的:

const appReducer = combineReducers({
  /* your app’s top-level reducers */
})

const rootReducer = (state, action) => {
  if (action.type === 'USER_LOGOUT') {
    return appReducer(undefined, action)
  }

  return appReducer(state, action)
}

如果使用redux-persist，可能还需要清理存储空间。Redux-persist将您的状态副本保存在存储引擎中，刷新时将从那里加载状态副本。

首先，您需要导入适当的存储引擎，然后在将其设置为undefined并清除每个存储状态键之前解析状态。

const rootReducer = (state, action) => {
    if (action.type === SIGNOUT_REQUEST) {
        // for all keys defined in your persistConfig(s)
        storage.removeItem('persist:root')
        // storage.removeItem('persist:otherKey')

        return appReducer(undefined, action);
    }
    return appReducer(state, action);
};

Dan Abramov的答案没有做的一件事是为参数化选择器清除缓存。如果你有一个这样的选择器:

export const selectCounter1 = (state: State) => state.counter1;
export const selectCounter2 = (state: State) => state.counter2;
export const selectTotal = createSelector(
  selectCounter1,
  selectCounter2,
  (counter1, counter2) => counter1 + counter2
);

然后你必须像这样在登出时释放它们:

selectTotal.release();

否则，最后一次调用选择器的记忆值和最后一个参数的值仍将在内存中。

代码示例来自ngrx文档。

只需将注销链接清除会话并刷新页面。您的商店不需要额外的代码。当您想要完全重置状态时，页面刷新是一种简单且易于重复的处理方法。

为了将状态重置为初始状态，我编写了以下代码:

const appReducers = (state, action) =>
   combineReducers({ reducer1, reducer2, user })(
     action.type === "LOGOUT" ? undefined : state,
     action
);

我已经创建了清除状态的操作。因此，当我分派登出动作创建者时，我也分派动作来清除状态。

用户记录动作

export const clearUserRecord = () => ({
  type: CLEAR_USER_RECORD
});

注销操作创建器

export const logoutUser = () => {
  return dispatch => {
    dispatch(requestLogout())
    dispatch(receiveLogout())
    localStorage.removeItem('auth_token')
    dispatch({ type: 'CLEAR_USER_RECORD' })
  }
};

减速机

const userRecords = (state = {isFetching: false,
  userRecord: [], message: ''}, action) => {
  switch (action.type) {
    case REQUEST_USER_RECORD:
    return { ...state,
      isFetching: true}
    case RECEIVE_USER_RECORD:
    return { ...state,
      isFetching: false,
      userRecord: action.user_record}
    case USER_RECORD_ERROR:
    return { ...state,
      isFetching: false,
      message: action.message}
    case CLEAR_USER_RECORD:
    return {...state,
      isFetching: false,
      message: '',
      userRecord: []}
    default:
      return state
  }
};

我不确定这是否是最佳的?