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。

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);
};

简单回答一下丹·阿布拉莫夫的问题:

const rootReducer = combineReducers({
    auth: authReducer,
    ...formReducers,
    routing
});


export default (state, action) =>
  rootReducer(action.type === 'USER_LOGOUT' ? undefined : state, action);

对我来说，一个快速而简单的选择是使用还原复位。这是简单的，也有一些高级选项，为较大的应用程序。

在创建存储区中设置

import reduxReset from 'redux-reset'
// ...
const enHanceCreateStore = compose(
    applyMiddleware(...),
    reduxReset()  // Will use 'RESET' as default action.type to trigger reset
)(createStore)
const store = enHanceCreateStore(reducers)

在注销函数中调度您的“重置”

store.dispatch({
    type: 'RESET'
})

Dan Abramov的答案是正确的，除了我们在使用react-router-redux包时遇到了一个奇怪的问题。

我们的解决方案是不将状态设置为undefined，而是仍然使用当前的路由减速器。因此，如果您正在使用这个包，我建议实现下面的解决方案

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

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
);