我不能在Redux reducer中使用state.filter

Question

所以我正在使用React-Redux，在我的reducer中我想從我的狀態中刪除一個主機名。 所以從環顧四周我發現state.filter是實現這一目標的好方法。 但是，這不起作用，我在控制台中不斷收到錯誤。

index.js

import "babel-polyfill";
import React from 'react';
import ReactDOM from 'react-dom';
import {Provider} from 'react-redux';
import thunkMiddleware  from 'redux-thunk';
import createLogger from 'redux-logger'
import { HostnameAppContainer } from './components/HostnameApp';
import {createStore, applyMiddleware} from 'redux';
import hostnameApp from './reducer'
import {fetchHostnames} from './action_creators'

const loggerMiddleware = createLogger()

const store = createStore(
    hostnameApp,
    applyMiddleware(
        thunkMiddleware, // lets us dispatch() functions
        loggerMiddleware // neat middleware that logs actions
    )
)

store.dispatch(fetchHostnames())

ReactDOM.render(
    <Provider store={store}>
        <HostnameAppContainer />
    </Provider>,
document.getElementById('app')
)
;

reducer.js

export default function(state = {hostnames:[]}, action) {
    switch (action.type) {
        case 'FETCH_HOSTNAMES':
            return Object.assign({}, state, {
                hostnames: action.hostnames
            })
        case 'DELETE_HOSTNAME':
            return state.filter(hostname =>
                hostname.id !== action.hostnameId
            )
        case 'ADDED_HOSTNAME':
            return Object.assign({}, state, {
                hostnames: [
                    ...state.hostnames,
                    action.object
                ]
            })
    }
    return state;
}

謝謝您提前獲取所有建議或解決方案。

Answer 1

在你的case 'DELETE_HOSTNAME'你正在調用state.filter 。 此時您的狀態將是Object而不是數組。

你可能想要做這樣的事情：

case 'DELETE_HOSTNAME':
  return { hostnames: state.hostnames.filter(hostname =>
     hostname.id !== action.hostnameId
  )}