How to open URL in flutter app with url_launcher? - flutter

Question

I use the package url_launcher to open url, here is the code:
Future <void> _openURL(String url) async{
if(await canLaunch(url)) {
await launch(
url,
forceSafariVC: false,
forceWebView: true,
);
}
else{
throw 'Cant open URL';
}
}
ListTile(
title: Text('Google'),
onTap: () {
_openURL('https://www.google.de/');
}),
But no matter what url i want to open, i get the error 'Cant open URL'

Answer 1

I get the same error: Unhandled Exception: Could not launch
As you can see here https://pub.dev/packages/url_launcher you have to put the following snippet at your AndroidManifest.xml
<queries>
<!-- If your app opens https URLs -->
<intent>
<action android:name="android.intent.action.VIEW" />
<data android:scheme="https" />
</intent>
</queries>

Answer 2

Try to below code:
Create launchURL function :
_launch() async {
const url = 'https://www.google.de/';
if (await canLaunch(url)) {
await launch(url);
} else {
throw 'Could not launch $url';
}
}
Create your Widget :
ListTile(
onTap: _launch,
title: Text('Google'),
),
Hope its solved your problem

Answer 3

I was not able to open the URL in Android using Flutter app with pub file https://pub.dev/packages/url_launcher
Since JavaScript was not enable.
Here's working code snippet for Android -
void _launchURL(String _url) async {
if (await canLaunch(_url)) {
await launch(_url, forceSafariVC: true, forceWebView: true, enableJavaScript: true);
} else {
throw 'Could not launch $_url';
}
}
Then I added a key enableJavaScript: true and it started working on Android too. iOS no need for configuration.