To open a browser on a specific given URL use: ShellExecute(self. WindowHandle,'open','www.cryer.co.uk',nil,nil, SW_SHOWNORMAL);
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',
},
});
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} />
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.
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