React Native 组件中的 onEnter/onExit 方法 (react-native-router-flux) [英] onEnter/onExit method in React Native Component (react-native-router-flux)
本文介绍了React Native 组件中的 onEnter/onExit 方法 (react-native-router-flux)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
所以我可以在路由器定义中我的应用程序的根目录中使用 onEnter/onExit 方法,它工作得非常好:
So I can use the onEnter/onExit method at the root of my app in the router definitions and it works perfectly fine:
<Scene key="arena" hideNavBar={true} onEnter={() => console.log("Entered")} component={ArenaPage} />
有什么办法可以在组件内部执行此操作,以便我可以更新本地状态??
Is there any way I can do this inside the component itself so that I can update the local state??
export default class ArenaPage extends Component {
onEnter () {
this.setState(...)
}
// Render blah blah blah...
}
如果不可能,是否有在离开场景导航时触发 componentWillUnmount (Actions.[key])
If not possible, Is there anyway to trigger componentWillUnmount when navigating away from Scene (Actions.[key])
推荐答案
您可以使用 onEnter
和 onExit
方法来获得您想要的结果.反过来可以像这样访问this
.
You can use onEnter
and onExit
methods to get the result you want. And in turn can access this
like so.
import { Actions } from 'react-native-router-flux'
class Home extends Component {
static onEnter() {
// homeSceneKey needs to be the same key that you use to configure the Scene
Actions.refs.homeSceneKey.getWrappedInstance().myFunction()
};
myFunction = () => {
// have access to this (props and state) here
this.blah = "what ever you want to do"
}
}
希望能帮到你
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