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
);

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;
  }
};

丹·阿布拉莫夫的回答帮我破案了。然而，我遇到了一个案例，并不是整个州都需要清理。所以我是这样做的:

const combinedReducer = combineReducers({
    // my reducers 
});

const rootReducer = (state, action) => {
    if (action.type === RESET_REDUX_STATE) {
        // clear everything but keep the stuff we want to be preserved ..
        delete state.something;
        delete state.anotherThing;
    }
    return combinedReducer(state, action);
}

export default rootReducer;

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。

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);
};

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)
}