firebase_messing如何清除通知

Question

我正在使用firebase_messaging当通知到达时，我将显示警报对话框。以下是我的代码。

showNotification(BuildContext context) {
    _firebaseMessaging.configure(
      onMessage: (Map<String, dynamic> message) async {
        print('on message $message');
        _showPushNotificationDialog(message['notification']['title'],
            message['notification']['body'], context);
      },
      onResume: (Map<String, dynamic> message) async {
        print('on resume $message');
        _showPushNotificationDialog(
            message['data']['title'], message['data']['body'], context);
      },
      onLaunch: (Map<String, dynamic> message) async {
        print('on launch $message');
        _showPushNotificationDialog(
            message['data']['title'], message['data']['body'], context);
      },
    );
  }

_showPushNotificationDialog方法在 onMessage ， onResume 和 onLaunch 方法方法call。

面对问题，例如我的应用程序处于后台或终止模式时，通知将来并利用通知托盘，都可以正常工作。但是，当我上另一页并回到上一个所有时间_firebaseMessaging.configure(....方法调用时，它都有数据，因此每次我的警报对话框pups up。

那么，如何清除通知托盘单击的通知？

Answer 1

我也有问题。我有这样的解决方法：我所做的是用静态布尔和静态方法创建一个类：

class MessagingWidget {
  static bool _isConfigured = false;

  static void configuringFirebase(User currentUser, BuildContext context){
      
      final FirebaseMessaging _firebaseMessaging = FirebaseMessaging();
      if (!_isConfigured) {
      _firebaseMessaging.configure(
      onMessage: (Map<String, dynamic> message) async {
        print("onMessage: $message");
        final notification = message['notification'];
      },
      onLaunch: (Map<String, dynamic> message) async {
        print("onLaunch: $message");
        final notification = message['data'];
        if(notification['title']!=null){
            if(notification['title']=="Testo"){
              goToAppointmentsScreen(currentUser,context);
            }
          }
      },
      onResume: (Map<String, dynamic> message) async {
        print("onResume: $message");
        final notification = message['data'];
        if(notification['title']!=null){
            if(notification['title']=="Testo"){
              goToAppointmentsScreen(currentUser,context);
            }
          }
      },
    );
    _isConfigured = true;
    }
  }
    
    
  }
  void goToAppointmentsScreen(User currentUser1, BuildContext context1) async {
    final bool backFromAppointmentsScreen=await Navigator.push(
            context1,
            MaterialPageRoute(builder: (context) => Appointment( 
              currentUser1),
            ),
            );
  }

然后，我从路由小部件的初始化中调用了此方法：

@override
  void initState(){
    super.initState();
    refreshServices();
    getDirectionBasedOnLocation();
    MessagingWidget.configuringFirebase(currentUser, context);
}

我希望这对您有帮助

Answer 2

我知道，这有点丑陋，但我只知道：

1. 添加flutter_local_notifications
2. import 'package:flutter_local_notifications/flutter_local_notifications.dart';

3. 只需创建final fln = FlutterLocalNotificationsPlugin();

4. 并在需要时使用fln.cancelAll()。

Answer 3

我为防止OnLaunch和OnRemume方法一次运行的方法是一次又一次运行的是使用最后一个通知(我使用shared_preferences(

检查当前通知

这是片段：

_firebaseMessaging.configure(
    onMessage: (Map<String, dynamic> message) async {
      print('on message $message');
      onMessageReceived(context, message);
    },
    onResume: (Map<String, dynamic> message) async {
      print('on resume $message');
      onLaunch(context, message);
    },
    onLaunch: (Map<String, dynamic> message) async {
      print('on launch $message');
      onLaunch(context, message);
    },
  );
.
.
.
void onLaunch(BuildContext context, Map<String, dynamic> remoteMessage) async {
  var pref = SharedPreferences.getInstance();
  var data = remoteMessage['data'] ?? remoteMessage;
  String lastData = '';
  await pref.then((prefs) {
    lastData = prefs.get('remote_message');
  });
  if ((data['type'] != null || data['id'] != null) &&
      data.toString() != lastData) {
    toDetailPageFromPush(
      context,
      data['type'],
      data['id'],
    );
    pref.then((prefs) {
      prefs.setString('remote_message', data.toString());
    });
  } else {
    print('on launch error $remoteMessage');
  }
}

Answer 4

这是另一个解决方案，使用 message_id 是通知中的唯一值，因此我们可以使用共享预处理保存通知的最后一个ID并比较带有当前通知：

processNotification(message, BuildContext context) async {
 try {
  SharedPreferences sharedPreferences = await SharedPreferences.getInstance();
 
  String lastMessageId = sharedPreferences.getString('last_message_id');
  String currentMessageId = message['data']['google.message_id'];
  //COMPARE NOTIFICATIONS ID
  if(currentMessageId != lastMessageId) {
    
     //SET LAST NOTIFICATION ID
     sharedPreferences.setString('last_message_id', currentMessageId);
 
     //SHOW A DIALOG OR NAVIGATE TO SOME VIEW, IN THIS CASE TO A VIEW
     String screen = message['data']['screen'];
     Navigator.of(context).pushNamed(screen);
  }
} catch (e) {
  print('ERROR PROCESSING NOTIFICATION');
  print(e);
 }
}

现在我们可以在配置中调用此功能：

 _firebaseMessaging.configure(
  onMessage: (Map<String, dynamic> message) async {
    print('on message $message');
    processNotification(context,message);
  },
  onResume: (Map<String, dynamic> message) async {
    print('on resume $message');
    processNotification(context,message);
  },
  onLaunch: (Map<String, dynamic> message) async {
    processNotification(context,message);
  },
);

Answer 5

尝试在initstate中而不是在自定义方法中配置firebasemessging。应该有效：(