Reactnative 在浏览器中打开链接并返回应用程序

我在react native中开发了一个应用程序，应该与支付网关通信，在完成支付过程(成功或失败)后，我需要向用户显示一个alert .为此，我在100中打开了一个链接，然后用101获得了return的url，并显示成功或失败消息.

但是，这security个问题的流程必须在default device browser中完成.

当前代码:

const BASEURL = 'https://gatewayURL/?ID=';
let Token = null;
let paymentAccepted = null;
let paymentFactorId = null;

class Gateway extends PureComponent {
  static propTypes = {
    dispatch: PropTypes.func,
    navigation: PropTypes.any,
  }

  componentWillMount() {
    this.props.dispatch(getPaymentStatus());
  }


  _onLoad(webViewState) {
    let url = webViewState.url.toString();
    let isResponseValid = url.includes('backFromGateway');
    if(isResponseValid){
      if(this.props.checkedPaymentStatus != 'checked' ){
        setTimeout(() => {
          this.props.dispatch(setPaymentStatus('checked'));

          let splitedURL = url.split("/");
          paymentFactorId = splitedURL[splitedURL.length -2];
          if(splitedURL[splitedURL.length - 1] === '0'){
            paymentAccepted = true;
            this.props.dispatch(setGatewayResponse('done', paymentFactorId));
          }
          else {
            paymentAccepted = false;
            this.props.dispatch(setGatewayResponse('rejected', paymentFactorId));
          }


          this.props.navigation.navigate('BackFromGateway', { title: '' })
        }, 1000);
      }
    }
  }


  render() {
    const { addNewOrderGatewayToken, checkedPaymentStatus } = this.props;
    token = addNewOrderGatewayToken;
    let view = null;
    if(checkedPaymentStatus !== 'checked'){
      view =  <WebView onNavigationStateChange={this._onLoad.bind(this)} style={styles.container} source={{ uri: `${BASEURL}${token}`  }}/>
    }
    else{
      view = <View></View>
    }

    return (
      <View style={styles.container}>
        {view}
      </View>      
    );
  }
}

知道吗