I want my flutter app run only if internet connection is available. If the internet is not present show a dialog(internet is not present) I'm using conectivity plugin but still not satisfied.
Here is my main function
Future main() async {
try {
final result = await InternetAddress.lookup('google.com');
if (result.isNotEmpty && result[0].rawAddress.isNotEmpty) {
print('connected');
}
} on SocketException catch (_) {
print('not connected');
}
runApp(MyApp());}
You can't use dialog in main() method directly because there is no valid context available yet.
Here is the basic code of what you are looking for.
void main() => runApp(MaterialApp(home: MyApp()));
class MyApp extends StatefulWidget {
@override
_MyAppState createState() => _MyAppState();
}
class _MyAppState extends State<MyApp> {
@override
void initState() {
super.initState();
Timer.run(() {
try {
InternetAddress.lookup('google.com').then((result) {
if (result.isNotEmpty && result[0].rawAddress.isNotEmpty) {
print('connected');
} else {
_showDialog(); // show dialog
}
}).catchError((error) {
_showDialog(); // show dialog
});
} on SocketException catch (_) {
_showDialog();
print('not connected'); // show dialog
}
});
}
void _showDialog() {
// dialog implementation
showDialog(
context: context,
builder: (context) => AlertDialog(
title: Text("Internet needed!"),
content: Text("You may want to exit the app here"),
actions: <Widget>[FlatButton(child: Text("EXIT"), onPressed: () {})],
),
);
}
@override
Widget build(BuildContext context) {
return Scaffold(
appBar: AppBar(title: Text("Internet")),
body: Center(
child: Text("Working ..."),
),
);
}
}
catchError() I have added it now and it is 100% working.