如何在 React Native App 中显示超链接？

像这样：

<Text style={{color: 'blue'}}
      onPress={() => Linking.openURL('http://google.com')}>
  Google
</Text>

使用与 React Native 捆绑在一起的模块。Linking

import { Linking } from 'react-native';

所选答案仅指 iOS。对于这两个平台，您可以使用以下组件：

import React, { Component, PropTypes } from 'react';
import {
  Linking,
  Text,
  StyleSheet
} from 'react-native';

export default class HyperLink extends Component {

  constructor(){
      super();
      this._goToURL = this._goToURL.bind(this);
  }

  static propTypes = {
    url: PropTypes.string.isRequired,
    title: PropTypes.string.isRequired,
  }

  render() {

    const { title} = this.props;

    return(
      <Text style={styles.title} onPress={this._goToURL}>
        >  {title}
      </Text>
    );
  }

  _goToURL() {
    const { url } = this.props;
    Linking.canOpenURL(url).then(supported => {
      if (supported) {
        Linking.openURL(this.props.url);
      } else {
        console.log('Don\'t know how to open URI: ' + this.props.url);
      }
    });
  }
}

const styles = StyleSheet.create({
  title: {
    color: '#acacac',
    fontWeight: 'bold'
  }
});