javascript - React Native - 如何在返回上一页时刷新 FlatList？

Question

我将尝试解释我的情况:

我有 2 个页面/屏幕

SreenA.js :

...
<FlatList>
</FlatList>
...

ScreenB.js :

...
<Button> This button for back to ScreenA </Button>
...

我想要实现的目标:

When i press back button in screenB , my app will back to screenA. When i come back to screenA , i need to refresh the FlatList. (each time i back to screenA).

我如何在 Android Native 中实现这种情况？

I add this section to help you to more understand what challenge i face. In Android Native, i will use onResume to refresh the list (onResume called everytime the screen/page called).

我已经尝试但现在对我不起作用的方法:

我已经尝试过了，但是当应用程序在后台然后在前台显示时，我得到的只是调用:

state = {
    appState: AppState.currentState
  }

  componentDidMount() {
    AppState.addEventListener('change', this._handleAppStateChange);
  }

  componentWillUnmount() {
    AppState.removeEventListener('change', this._handleAppStateChange);
  }

  _handleAppStateChange = (nextAppState) => {
    if (this.state.appState.match(/inactive|background/) && nextAppState === 'active') {
      console.log('App has come to the foreground!')
      //i place the code here for refresh the flatList. But not work
    }
    this.setState({appState: nextAppState});
  }

Answer 1

添加 navigation.addListener 以在每次页面接收到导航焦点时触发。

举个例子:

class Page1 extends React.Component {
  constructor(props) {
    super(props);
    this.state = { ts: 0 }

    // this will fire every time Page 1 receives navigation focus
    this.props.navigation.addListener('willFocus', () => {
      this.setState({ ts: Date.now() })
    })
  }

  render() {
    return (
      <View style={styles.container}>
        <Text style={styles.paragraph}>
          Page 1 [{this.state.ts}]
        </Text>
        <Button title="Page 2" onPress={() => this.props.navigation.navigate('Page2')}/>
      </View>
    );
  }
}

有关完整示例，请查看以下 snack在你的手机上。每次您导航回第 1 页时，时间戳都会更新，但不会在应用恢复时更新。

您应该能够根据您的需要从这个示例中扩展它。