单击通知按钮不会在颤动中打开应用程序

Posted 2023-02-23

【中文标题】单击通知按钮不会在颤动中打开应用程序

【英文标题】：Clicking on Notification button not opening app in flutter

【发布时间】：2020-02-01 15:44:12

【问题描述】：

我已经集成了用于通知的 firebase-messaging 插件，但是在单击通知时，如果它在后台或被杀死，则应用程序不会打开。我想在单击 natification 时打开应用程序。

下面是我的代码

    void firebaseCloudMessagingListeners() 
      FirebaseMessaging _firebaseMessaging = new FirebaseMessaging();
      _firebaseMessaging.getToken().then((token)
        print(token);
      );

     _firebaseMessaging.requestNotificationPermissions(
            const iosNotificationSettings(sound: true, badge: true, alert: true));
        _firebaseMessaging.onIosSettingsRegistered
            .listen((IosNotificationSettings settings) 
          print("Settings registered: $settings");
        );

      _firebaseMessaging.configure(
        onMessage: (Map<String, dynamic> message) async 
          _showNotificationWithDefaultSound(message['notification']['body']);

          print('on message $message');
          return;
        ,
        onResume: (Map<String, dynamic> message) async 
          _showNotificationWithDefaultSound(message['data']['body']);
          print('on message $message');
          return;
        ,

        onLaunch: (Map<String, dynamic> message) async 
          _showNotificationWithDefaultSound(message['data']['body']);
          print('on message $message');
        ,
      );
    

    _showNotificationWithDefaultSound(message) async 
      FlutterLocalNotificationsPlugin flutterLocalNotificationsPlugin =
          new FlutterLocalNotificationsPlugin();

      var android = new AndroidInitializationSettings('@mipmap/ic_launcher');
      var ios = new IOSInitializationSettings();
      var initSettings = new InitializationSettings(android, ios);
      flutterLocalNotificationsPlugin.initialize(initSettings,
          onSelectNotification: onSelectNotification);

      var android1 = new AndroidNotificationDetails(
          'channel id', 'channel NAME', 'channel DESCRIPTION',
          importance: Importance.Max, priority: Priority.High, ticker: 'ticker');

      var ios1 = new IOSNotificationDetails();
      var platform = new NotificationDetails(android1, ios1);
      await flutterLocalNotificationsPlugin
          .show(0, message, 'App Notification!', platform, payload: message);
    

【参考方案1】：

你需要添加AndroidManifest.xml

  <intent-filter>
      <action android:name="FLUTTER_NOTIFICATION_CLICK" />
      <category android:name="android.intent.category.DEFAULT" />
  </intent-filter>

然后，您需要在 firebase 通知的数据正文中发送 click_action:FLUTTER_NOTIFICATION_CLICK，如下所示：

  DATA='
  "notification": 
    "body": "this is a body",
    "title": "this is a title"        
  ,
  "data": 
    "att1": "value..", 
    "att2": "value..",        
    "click_action": "FLUTTER_NOTIFICATION_CLICK",
  ,
  "to": "<FCM TOKEN>"
'

当你的应用被杀死时，如果你点击通知，就会调用onLaunch方法，你必须通过message['data']获取参数。

在此处查看更多信息：https://***.com/a/48405551/7105694

【参考方案2】：

可能你需要实现onSelectNotification()方法

Future<void> onSelectNotification(String payload) async 
        Navigator.push(
            context, PageTransition(
            type: PageTransitionType.leftToRight,
            child: /* your home screen name */));
  

【讨论】：

但是如果应用程序被杀死会怎样