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);
};

如果你想重置一个减速机

例如

const initialState = { isLogged:假 ｝ //这将是你的行动 export const resetReducer = () => { 返回{ 类型:“重置” ｝ ｝ 导出默认值(state = initialState， { 类型, 有效载荷 }) => { 开关(类型){ //你的行动会报应到她头上 例“重置”: 返回{ initialState… ｝ ｝ ｝ //从你的前端 调度(resetReducer ())

在服务器中，有一个变量:global。isSsr = true 在每个reducer中，我都有一个const: initialState 要重置存储中的数据，我对每个Reducer执行以下操作:

appReducer.js的例子:

 const initialState = {
    auth: {},
    theme: {},
    sidebar: {},
    lsFanpage: {},
    lsChatApp: {},
    appSelected: {},
};

export default function (state = initialState, action) {
    if (typeof isSsr!=="undefined" && isSsr) { //<== using global.isSsr = true
        state = {...initialState};//<= important "will reset the data every time there is a request from the client to the server"
    }
    switch (action.type) {
        //...other code case here
        default: {
            return state;
        }
    }
}

最后在服务器的路由器上:

router.get('*', (req, res) => {
        store.dispatch({type:'reset-all-blabla'});//<= unlike any action.type // i use Math.random()
        // code ....render ssr here
});

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。

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
);