Flutter Awesome Notification couldn't bind a method to action button
2,221
You should create a event stream and listen to events like that:
AwesomeNotifications().actionStream.listen((event) {
print('event received!');
print(event.toMap().toString());
// do something based on event...
});
Author by
arincgurkan
Updated on December 01, 2022Comments
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arincgurkan over 1 year
I'm developing an app which needs to answer the notifications. I can display the button when I received the notification. However, I couldn't bind any method to it. Here is my code:
Future<void> firebaseMessagingBackgroundHandler(RemoteMessage message) async { await Firebase.initializeApp(); AwesomeNotifications().createNotification( content: NotificationContent( id: 10, channelKey: 'basic_channel', title: 'Simple Notification', body: 'Simple body'), actionButtons: [ NotificationActionButton( label: 'TEST', enabled: true, buttonType: ActionButtonType.Default, key: 'test', ) ]); print("Background message service"); }
Thank you for your helps!
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Benyamin Beyzaie almost 3 yearsany solution on that?
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alperefesahin over 1 yearoutdated for the new version...