onLogout () { this.props.history.push(' /登录');//发送用户到登录页面 window.location.reload ();//刷新页面 ｝

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 {}

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)
}

NGRX4更新

如果您正在迁移到NGRX 4，您可能已经从迁移指南中注意到用于组合reducer的rootreducer方法已经被ActionReducerMap方法所取代。起初，这种新的做事方式可能会使重置状态成为一个挑战。它实际上很简单，但这样做的方式已经改变了。

这个解决方案的灵感来自NGRX4 Github文档的元还原器API部分。

首先，让我们假设你正在使用NGRX的新ActionReducerMap选项像这样组合你的reducer:

//index.reducer.ts
export const reducers: ActionReducerMap<State> = {
    auth: fromAuth.reducer,
    layout: fromLayout.reducer,
    users: fromUsers.reducer,
    networks: fromNetworks.reducer,
    routingDisplay: fromRoutingDisplay.reducer,
    routing: fromRouting.reducer,
    routes: fromRoutes.reducer,
    routesFilter: fromRoutesFilter.reducer,
    params: fromParams.reducer
}

现在，假设你想从app。module内部重置状态

//app.module.ts
import { IndexReducer } from './index.reducer';
import { StoreModule, ActionReducer, MetaReducer } from '@ngrx/store';
...
export function debug(reducer: ActionReducer<any>): ActionReducer<any> {
    return function(state, action) {

      switch (action.type) {
          case fromAuth.LOGOUT:
            console.log("logout action");
            state = undefined;
      }
  
      return reducer(state, action);
    }
  }

  export const metaReducers: MetaReducer<any>[] = [debug];

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

export class AppModule { }

这基本上是用NGRX 4达到同样效果的一种方法。

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
  })),