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;
  }
};

一种方法是在应用程序中编写一个根减速器。

根减速机通常会将处理操作委托给combineReducers()生成的减速机。但是，无论何时它接收到USER_LOGOUT操作，它都会再次返回初始状态。

例如，如果你的根减速器是这样的:

const rootReducer = combineReducers({
  /* your app’s top-level reducers */
})

你可以将它重命名为appReducer，并编写一个新的rootReducer委托给它:

const appReducer = combineReducers({
  /* your app’s top-level reducers */
})

const rootReducer = (state, action) => {
  return appReducer(state, action)
}

现在我们只需要教新的rootReducer返回初始状态以响应USER_LOGOUT操作。如我们所知，无论操作如何，当调用以undefined作为第一个参数时，约简器都应该返回初始状态。让我们用这个事实来有条件地剥离累积状态，当我们把它传递给appReducer:

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

  return appReducer(state, action)
}

现在，每当USER_LOGOUT触发时，所有减约器都将重新初始化。它们还可以返回与初始值不同的值，因为它们可以检查动作。也要打字。

重申一下，完整的新代码是这样的:

const appReducer = combineReducers({
  /* your app’s top-level reducers */
})

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

  return appReducer(state, action)
}

如果使用redux-persist，可能还需要清理存储空间。Redux-persist将您的状态副本保存在存储引擎中，刷新时将从那里加载状态副本。

首先，您需要导入适当的存储引擎，然后在将其设置为undefined并清除每个存储状态键之前解析状态。

const rootReducer = (state, action) => {
    if (action.type === SIGNOUT_REQUEST) {
        // for all keys defined in your persistConfig(s)
        storage.removeItem('persist:root')
        // storage.removeItem('persist:otherKey')

        return appReducer(undefined, action);
    }
    return appReducer(state, action);
};

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。

我已经创建了一个组件来给Redux重置状态的能力，你只需要使用这个组件来增强你的存储和分派一个特定的动作。类型触发重置。执行的想法和Dan Abramov在他们的回答中说的是一样的。

Github: https://github.com/wwayne/redux-reset

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)
}

使用Redux Toolkit和/或Typescript:

const appReducer = combineReducers({
  /* your app’s top-level reducers */
});

const rootReducer = (
  state: ReturnType<typeof appReducer>,
  action: AnyAction
) => {
/* if you are using RTK, you can import your action and use it's type property instead of the literal definition of the action  */
  if (action.type === logout.type) {
    return appReducer(undefined, { type: undefined });
  }

  return appReducer(state, action);
};