redux - 错误 : Reducer "counter" returned undefined during initialization. 如果传递给 reducer 的状态未定义

Question

我正在尝试对 reduxer(Learning) 进行基本测试，它看起来很基本，没有问题，但我得到了这个错误"×错误: reducer “计数器”在初始化期间返回未定义。如果传递给 reducer 的状态未定义，则必须显式返回初始状态。初始状态可能不是未定义的。如果你不想为这个 reducer 设置一个值，你可以使用 null 而不是 undefined。”

我认为这种回应是非常愚蠢的希望得到帮助。

Reduxer 文件。计数器.js

const counterReducer = (state = 0, action) => {
  switch (action.type) {
    case "INCREMENT":
      return state + 1;
    case "DECREMENT":
      return state - 1;
  }
};

export default counterReducer;

isLogged.js

const loggedReducer = (state = false, action) => {
  switch (action.type) {
    case "SING_IN":
      return !state;
    default:
      return state;
  }
};
export default loggedReducer;

index.js

import counterReducer from "./counter";
import loggedReducer from "./isLogged";
import { combineReducers } from "redux";

const allReducers = combineReducers({
  counter: counterReducer,
  isLogged: loggedReducer,
});

export default allReducers;

src 文件夹>index.js.

import React from "react";
import ReactDOM from "react-dom";
import "./index.css";
import App from "./App";
import * as serviceWorker from "./serviceWorker";
import { createStore } from "redux";
import allReducer from "./reducers";

const store = createStore(
  allReducer,
  window.__REDUX_DEVTOOLS_EXTENSION__ && window.__REDUX_DEVTOOLS_EXTENSION__()
);

ReactDOM.render(
  <React.StrictMode>
    <App />
  </React.StrictMode>,
  document.getElementById("root")
);


serviceWorker.unregister();

Answer 1

你应该默认从reducer返回

const counterReducer = (state = 0, action) => {
  switch (action.type) {
    case "INCREMENT":
      return state + 1;
    case "DECREMENT":
      return state - 1;
    default:
      return state;
  }
};

每个 reducer 都会收到分派(dispatch)的每个 action。如果您不知道操作类型，您应该返回原样的状态。