定义一个动作:

const RESET_ACTION = {
  type: "RESET"
}

然后在每个减速器中假设您使用switch或if-else通过每个减速器处理多个动作。我要拿开关的箱子。

const INITIAL_STATE = {
  loggedIn: true
}

const randomReducer = (state=INITIAL_STATE, action) {
  switch(action.type) {
    case 'SOME_ACTION_TYPE':

       //do something with it

    case "RESET":

      return INITIAL_STATE; //Always return the initial state

   default: 
      return state; 
  }
}

这样当你调用RESET动作时，你的reducer就会用默认状态更新存储。

现在，注销，你可以处理如下:

const logoutHandler = () => {
    store.dispatch(RESET_ACTION)
    // Also the custom logic like for the rest of the logout handler
}

每次用户登录时，不需要刷新浏览器。Store将始终处于默认状态。

store.dispatch(RESET_ACTION)只是详细阐述了这个思想。你很可能会有一个用于此目的的动作创建者。更好的方法是使用LOGOUT_ACTION。

一旦分派了这个LOGOUT_ACTION。然后，一个自定义中间件可以用Redux-Saga或Redux-Thunk拦截这个动作。然而，这两种方法都可以分派另一个动作“RESET”。通过这种方式，存储注销和重置将同步发生，您的存储将为另一个用户登录做好准备。

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);

From a security perspective, the safest thing to do when logging a user out is to reset all persistent state (e.x. cookies, localStorage, IndexedDB, Web SQL, etc) and do a hard refresh of the page using window.location.reload(). It's possible a sloppy developer accidentally or intentionally stored some sensitive data on window, in the DOM, etc. Blowing away all persistent state and refreshing the browser is the only way to guarantee no information from the previous user is leaked to the next user.

(当然，作为共享计算机上的用户，你应该使用“私人浏览”模式，自己关闭浏览器窗口，使用“清除浏览数据”功能，等等，但作为开发人员，我们不能期望每个人都总是那么勤奋)

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)
}

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);
};

下面的解决方案对我很有效。

我添加了重置状态功能到元还原器。关键在于使用

return reducer(undefined, action);

将所有减速器设置为初始状态。返回undefined反而会导致错误，因为存储的结构已经被破坏。

/还原剂index.ts

export function resetState(reducer: ActionReducer<State>): ActionReducer<State> {
  return function (state: State, action: Action): State {

    switch (action.type) {
      case AuthActionTypes.Logout: {
        return reducer(undefined, action);
      }
      default: {
        return reducer(state, action);
      }
    }
  };
}

export const metaReducers: MetaReducer<State>[] = [ resetState ];

app.module.ts

import { StoreModule } from '@ngrx/store';
import { metaReducers, reducers } from './reducers';

@NgModule({
  imports: [
    StoreModule.forRoot(reducers, { metaReducers })
  ]
})
export class AppModule {}