[android] Open Url in default web browser

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

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

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

In React 16.8+, using functional components, you would do

import React from 'react';
import { Button, Linking } from 'react-native';

const ExternalLinkBtn = (props) => {
  return <Button
            title={props.title}
            onPress={() => {
                Linking.openURL(props.url)
                .catch(err => {
                    console.error("Failed opening page because: ", err)
                    alert('Failed to open page')
                })}}
          />
}

export default function exampleUse() {
  return (
    <View>
      <ExternalLinkBtn title="Example Link" url="https://example.com" />
    </View>
  )
}

Try this:

import React, { useCallback } from "react";
import { Linking } from "react-native";
OpenWEB = () => {
  Linking.openURL(url);
};

const App = () => {
  return <View onPress={() => OpenWeb}>OPEN YOUR WEB</View>;
};

Hope this will solve your problem.

A simpler way which eliminates checking if the app can open the url.

  loadInBrowser = () => {
    Linking.openURL(this.state.url).catch(err => console.error("Couldn't load page", err));
  };

Calling it with a button.

<Button title="Open in Browser" onPress={this.loadInBrowser} />