在默认 Web 浏览器中打开 Url

问题描述

我是 react-native 的新手，我想在 默认浏览器(如 Android 和 iPhone 中的 Chrome)中打开 url.

I am new in react-native and i want to open url in default browser like Chrome in Android and iPhone both.

我们通过 Android 中的意图打开 url 与我想要实现的功能相同.

We open url via intent in Android same like functionality i want to achieve.

我已经搜索了很多次，但它会给我 Deepklinking 的结果.

I have search many times but it will give me the result of Deepklinking.

推荐答案

你应该使用 链接.

You should use Linking.

文档中的示例:

class OpenURLButton extends React.Component {
  static propTypes = { url: React.PropTypes.string };
  handleClick = () => {
    Linking.canOpenURL(this.props.url).then(supported => {
      if (supported) {
        Linking.openURL(this.props.url);
      } else {
        console.log("Don't know how to open URI: " + this.props.url);
      }
    });
  };
  render() {
    return (
      <TouchableOpacity onPress={this.handleClick}>
        {" "}
        <View style={styles.button}>
          {" "}<Text style={styles.text}>Open {this.props.url}</Text>{" "}
        </View>
        {" "}
      </TouchableOpacity>
    );
  }
}

这里有一个例子，你可以试试Expo Snack:

Here's an example you can try on Expo Snack:

import React, { Component } from 'react';
import { View, StyleSheet, Button, Linking } from 'react-native';
import { Constants } from 'expo';

export default class App extends Component {
  render() {
    return (
      <View style={styles.container}>
       <Button title="Click me" onPress={ ()=>{ Linking.openURL('https://google.com')}} />
      </View>
    );
  }
}

const styles = StyleSheet.create({
  container: {
    flex: 1,
    alignItems: 'center',
    justifyContent: 'center',
    paddingTop: Constants.statusBarHeight,
    backgroundColor: '#ecf0f1',
  },
});

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