How do you catch canLaunch exceptions using url_launcher in flutter? - flutter

Question

Using the code from the package I was unable to catch the exception. Note that I would like to catch this specific exception.
// from https://pub.dev/packages/url_launcher
_launchURL() async {
const url = 'myscheme://myurl';
if (await canLaunch(url)) {
await launch(url);
} else {
throw 'Could not launch $url';
}
}
// my code
try {
_launchURL();
}
catch (e)
{
// although the exception occurs, this never happens, and I would rather catch the exact canLaunch exception
}

Answer 1

I would try to put the try catch statement inside the function. I believe what is happening is that the try/catch statement is only applying for the function call and although it is async I dont believe that it actually tries and returns exeptions.
So the solution would look somethink like this:
_launchURL() async {
try{
const url = 'myscheme://myurl';
if (await canLaunch(url)) {
await launch(url);
} else {
throw 'Could not launch $url';
}
}
catch(e){
//debug e
}
}
// my code
launchURL();

Answer 2

You can use .then() for business logic.
For me, it is used to check if the app can be opened on the device.
Can be solution below,
--> url_launcher: ^6.0.2
--> https://pub.dev/packages/url_launcher
launch(appLink).then(
(bool isLaunch) {
print('isLaunch: $isLaunch');
if (isLaunch) {
// Launch Success
} else {
// Launch Fail
}
},
onError: (e) {
print('onError: $e');
},
).catchError(
(ex) => print('catchError: $ex'),
);
Work for me.

Answer 3

Future<void> _launch(String url) async {
await canLaunch(url)
? await launch(url)
: throw 'Could not launch $url';
}