java - 如果推送通知包含特定元素，则隐藏它

Question

如果有例如键 update，我不会显示从出现的顶部通知菜单收到的推送通知我的通知。现在，如果我收到带有此键的通知，所有通知都在通知栏中。我不想向用户显示此通知。

我正在使用 WakefulBroadcastReceiver 来处理如下通知:

public class PusherReceiver extends WakefulBroadcastReceiver {
    private boolean isAppOnForeground(Context context) {
        ActivityManager activityManager = (ActivityManager) context.getSystemService(Context.ACTIVITY_SERVICE);
        List<ActivityManager.RunningAppProcessInfo> appProcesses = activityManager.getRunningAppProcesses();
        if (appProcesses == null) 
            return false;

        final String packageName = context.getPackageName();
        for (ActivityManager.RunningAppProcessInfo appProcess : appProcesses) {
            if (appProcess.importance == ActivityManager.RunningAppProcessInfo.IMPORTANCE_FOREGROUND && appProcess.processName.equals(packageName)) {
                return true;
            }
        }

        return false;
    }

    public void onReceive(final Context context, Intent intent) {
        Log.i("SimpleWakefulReceiver", "Starting service @ " + SystemClock.elapsedRealtime());
        if (!isAppOnForeground((context))) {
            String pushNotificationBody = intent.getStringExtra("alert");

            try {
                JSONObject notificationData = new JSONObject(pushNotificationBody);

                // This is the Intent to deliver to our service.
                Intent service = new Intent(context, BackgroundService.class);
                // Put here your data from the json as extra in in the intent
                service.putExtra("notification", pushNotificationBody);

                Log.i("PUSH_NOTIFICATION_JSON", "RECEIVED JSON " + notificationData);

                // Start the service, keeping the device awake while it is launching.
                if (!notificationData.has("update")) {
                    startWakefulService(context, service);
                } else {
                    // Do nothing
                }
            } catch (JSONException e) {
                e.printStackTrace();
            }
        }
    }
}

更新:

我稍微改变了项目，使用 Onesignal 和他的 NotificationExtenderService，我做了如下的事情:

public class NotificationNotDisplayingExtender extends NotificationExtenderService {
    @Override
    protected boolean onNotificationProcessing(OSNotificationReceivedResult receivedResult) {
        String notification = receivedResult.toString();
        String notificationBody = receivedResult.payload.body;
        JSONObject notificationBodyJSON = null;
        try {
            notificationBodyJSON = new JSONObject(notificationBody);
        } catch (JSONException e) {
            e.printStackTrace();
        }
        JSONObject pushNotificationData = notificationBodyJSON;
        boolean hidden = false;

        if (pushNotificationData.has("update")) {
            Log.i("NOTIFICATION MANAGER", "PREVENT DISPLAY NOTIFICATION");

            hidden = true;
        }


        // Return true to stop the notification from displaying.
        return hidden;
    }
}

它阻止显示带有更新 key 的通知，但现在我没有在我的 PusherReceiver 中收到它来启动我的服务。有没有简单的方法将数据从我的 NotificationNotDisplayingExtender receivedResult 发送到我的 PusherReceiver？

现在看来我的 PusherReceiver 没有触发他的 onReceive 方法。

非常感谢您的提前帮助。

Answer 1

有两种类型的负载。 1.数据 2.通知

https://developers.google.com/cloud-messaging/concept-options

仅使用数据负载。然后你总是在 FirebaseMessagingService 中调用 onMessageRececived 方法