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.
You should use Linking.
Example from the docs:
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>
);
}
}
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',
},
});
.html at the end of the URL makes the link work again. Updated my answer, thanks for the heads up.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} />
openURL method. For example: If (this.state.url) Linking.openURL(this.state.url). You can also put the catch clause to use if you don't want to check prior.
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.
In React 16.8+, the following can be used to create an ExternalLinkBtn component, similar to an a tag on the web. The link will be opened externally in the default browser.
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')
})}}
/>
}
Below is an example of using our ExternalLinkBtn in a component
export default function exampleUse() {
return (
<View>
<ExternalLinkBtn title="Example Link" url="https://example.com" />
</View>
)
}
