react-native componentWillUnmount 在导航时不起作用

Question

我正在做这个简单的步骤，但 unmount 没有打电话，我不知道为什么。请我需要一个解决方案，我需要卸载才能在导航到另一个屏幕时被调用...

class Homemain extends Component {
    constructor(props) {
        super(props);
    }

    componentWillMount(){
        alert('willMount')
    }
    componentDidMount(){
        alert('didMount')
    }
    componentWillUnmount(){
        alert('unMount')
    }
    Details = () => {
        this.props.navigation.navigate('routedetailsheader')
    }

    render() {
        return(
            <View style={styles.container}>
                <TouchableOpacity onPress={() => this.Details()} style={{ flex: .45, justifyContent: 'center', alignItems: 'center', marginTop: '10%', marginRight: '10%' }}>
                    <Image
                        source={require('./../Asset/Images/child-notification.png')}
                        style={{ flex: 1, height: height / 100 * 20, width: width / 100 * 20, resizeMode: 'contain' }} />
                    <Text
                        style={{ flex: 0.5, justifyContent: 'center', fontSize: width / 100 * 4, fontStyle: 'italic', fontWeight: '400', color: '#000', paddingTop: 10 }}>Details</Text>
                </TouchableOpacity>
            </View>
        );
    }
}
export default (Homemain);

这是我的 RouteConfiguration，我将导航到下一个屏幕。有人可以帮我解决这个错误，以便我可以继续下一步

import React, { Component } from 'react';
import PropTypes from 'prop-types';
import { addNavigationHelpers, NavigationActions } from 'react-navigation';
import { connect } from 'react-redux';
import { BackHandler } from 'react-native';
import { Stack } from './navigationConfiguration';

const getCurrentScreen = (navigationState) => {
  if (!navigationState) {
    return null
  }
  const route = navigationState.routes[navigationState.index]
  if (route.routes) {
    return getCurrentScreen(route)
  }
  return route.routeName
}
class StackNavigation extends Component {
  static propTypes = {
    dispatch: PropTypes.func.isRequired,
    navigation: PropTypes.shape().isRequired,
  };

  constructor(props) {
    super(props);
    BackHandler.addEventListener('hardwareBackPress', this.backAction);
  }

  //backAction = () => this.navigator.props.navigation.goBack();

  backAction = () => {
    const { dispatch, navigation } = this.props;
    const currentScreen = getCurrentScreen(navigation)

    if (currentScreen === 'Homemain') {
      return false
    }
    else
      if (currentScreen === 'Login') {
        return false
      }

    dispatch(NavigationActions.back());
    return true;
  };

  render() {
    const { dispatch, navigation } = this.props;

    return (
      <Stack
        ref={(ref) => { this.navigator = ref; }}
        navigation={
          addNavigationHelpers({
            dispatch,
            state: navigation,
          })
        }
      />
    );
  }
}

export default connect(state => ({ navigation: state.stack }))(StackNavigation);

Answer 1

路由到新屏幕不会卸载当前屏幕。

对于您的用例，您无需在 componentWillUnmount 中编写代码，而是在 Details 本身中调用导航后继续编写代码。

如果您在从新屏幕按返回键返回当前屏幕时正在寻找回调。使用 goBack 如 https://github.com/react-navigation/react-navigation/issues/733 所示