React Native中如何在Redux action内实现登录成功后跳转至Account页面
可行解决方案
方案1:Login组件监听登录状态跳转(最推荐,耦合度最低)
该方案不需要把导航实例传入redux action层,符合分层逻辑,适配所有react-navigation版本:
- 先在
actions/user.js的登录thunk中,登录成功后dispatch action将登录状态、用户信息存入redux store:
// actions/user.js export const loginAction = (loginForm) => async (dispatch) => { try { const userData = await loginApi(loginForm) // 登录成功更新store状态 dispatch({ type: 'LOGIN_SUCCESS', payload: userData }) } catch(e) { dispatch({ type: 'LOGIN_FAIL', payload: e.message }) } }
- 在
Login.js中订阅store的登录状态,状态变更时触发跳转:
// screens/Login.js import { useSelector, useDispatch } from 'react-redux' import { useEffect } from 'react' import { loginAction } from '../actions/user' const Login = ({ navigation }) => { const dispatch = useDispatch() // 订阅store中的登录状态 const { isLoginSuccess } = useSelector(state => state.userModule) const handleLoginSubmit = () => { dispatch(loginAction({ username: 'xxx', password: 'xxx' })) } // 监听登录状态,成功后跳转 useEffect(() => { if(isLoginSuccess) { navigation.navigate('Account') // 可选:重置登录状态,避免返回Login页时重复触发跳转 dispatch({ type: 'RESET_LOGIN_STATUS' }) } }, [isLoginSuccess, navigation, dispatch]) // 其余页面渲染逻辑 }
方案2:封装全局导航服务(适配非组件场景跳转需求)
如果确实需要在action层直接调用跳转,可以封装全局导航ref,不需要传递navigation参数:
- 在
navigation目录下新增RootNavigation.js:
// navigation/RootNavigation.js import { createNavigationContainerRef } from '@react-navigation/native' export const navigationRef = createNavigationContainerRef() export function globalNavigate(routeName, params) { if (navigationRef.isReady()) { navigationRef.navigate(routeName, params) } }
- 给App入口的NavigationContainer绑定ref:
// App.js import { NavigationContainer } from '@react-navigation/native' import { navigationRef } from './navigation/RootNavigation' import TabNavigator from './navigation/tabNavigator' export default function App() { return ( <NavigationContainer ref={navigationRef}> <TabNavigator /> </NavigationContainer> ) }
- 任意位置(包括thunk中)都可以直接调用跳转:
// actions/user.js import { globalNavigate } from '../navigation/RootNavigation' export const loginAction = (loginForm) => async (dispatch) => { try { const userData = await loginApi(loginForm) dispatch({ type: 'LOGIN_SUCCESS', payload: userData }) // 直接调用全局跳转方法 globalNavigate('Account') } catch(e) { dispatch({ type: 'LOGIN_FAIL', payload: e.message }) } }
通用排查点
- 确认
Account路由已经正确注册到导航器中,路由名字母大小写完全匹配 - 如果是嵌套导航,跳转时需要指定父路由:
navigation.navigate('TabHome', { screen: 'Account' }) - 之前传navigation参数的方案失效,优先检查调用thunk时是否正确把navigation传入函数参数
内容的提问来源于stack exchange,提问作者Pruteanu Alexandru
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