Flutter - Run function after showDialog is dismissed

showDialog() can await a callback, and Navigator.pop can pass a value back. so instead of:

Future<Null> gewinner(int gewinner_team, List<String> spieler){
   return showDialog(
       ....
   );
}

you can use:

Future<Null> gewinner(int gewinner_team, List<String> spieler) async {
   String returnVal = await showDialog(
       ....
   );
}

and then in the dialog builder/screen you simply pop with a return value:

Navigator.pop(context, 'success');

and then do with the returnVal what you wish.

if (returnVal == 'success') {
  ...
}

if the dialog is dismissed then returnVal will be null.

Here is the simple example of getting status from Alert Dialog

    RaisedButton(
            onPressed: () {
              showDialog(
                  context: context,
                  builder: (context) => AlertDialog(
                        title: Text('Are you sure?'),
                        content: Text('Do you want to remove item?'),
                        actions: <Widget>[
                          FlatButton(
                              onPressed: () => Navigator.of(context).pop('Success'),
                              child: Text('NO')),
                          FlatButton(
                              onPressed: () => Navigator.of(context).pop('Failure'),
                              child: Text('YES'))
                        ],
                      )).then((value) =>
                  print('Result: ' + value.toString()));
            },
            child: Text('Show Alert Dialog'),
          ),