Question

我正在使用Parse.com的推送通知服务。根据{{3}}：

覆盖onPushReceive以触发“无声”的后台操作推

我找到了onPushOpen（）doc的源代码，但现在我必须重写onPushReceive（）来自定义声音和振动的行为。我不知道我应该在onPushReceive（）中做什么，是否有任何示例代码可以帮助我找出onPushReceive（）中的逻辑？感谢。

Answer 1

创建一个扩展ParsePushBroadcastReceiver的新类：

public class MyPushBroadcastReceiver extends ParsePushBroadcastReceiver {

public static final String PARSE_DATA_KEY = "com.parse.Data";

   @Override
   protected Notification getNotification(Context context, Intent intent) {
      // deactivate standard notification
      return null;
   }

   @Override
   protected void onPushOpen(Context context, Intent intent) {
      // Implement       
   }  

   @Override
   protected void onPushReceive(Context context, Intent intent) {
      JSONObject data = getDataFromIntent(intent);
      // Do something with the data. To create a notification do:

      NotificationManager notificationManager =
      (NotificationManager) context.getSystemService(Context.NOTIFICATION_SERVICE);

      NotificationCompat.Builder builder = new NotificationCompat.Builder(context);
      builder.setContentTitle("Title");
      builder.setContentText("Text");
      builder.setSmallIcon(R.drawable.ic_notification);
      builder.setAutoCancel(true);

      // OPTIONAL create soundUri and set sound:
      builder.setSound(soundUri);

      notificationManager.notify("MyTag", 0, builder.build());

   }

   private JSONObject getDataFromIntent(Intent intent) {
      JSONObject data = null;
      try {
         data = new JSONObject(intent.getExtras().getString(PARSE_DATA_KEY));
      } catch (JSONException e) {
         // Json was not readable...
      }
      return data;
   }
}

在你的Manifest中添加：

  <receiver
     android:name=".MyPushBroadcastReceiver"
     android:exported="false">
     <intent-filter>
        <action android:name="com.parse.push.intent.RECEIVE" />
        <action android:name="com.parse.push.intent.DELETE" />
        <action android:name="com.parse.push.intent.OPEN" />
     </intent-filter>
  </receiver>

更多信息：http://www.androidhive.info/2015/06/android-push-notifications-using-parse-com/

Answer 2

您可以使用intent额外参数“action”来调用您的意图来处理您想要的任何内容。

原创onPushReceive来源：

protected void onPushReceive(Context context, Intent intent) {
    JSONObject pushData = null;

    try {
        pushData = new JSONObject(intent.getStringExtra("com.parse.Data"));
    } catch (JSONException var7) {
        Parse.logE("com.parse.ParsePushReceiver", "Unexpected JSONException when receiving push data: ", var7);
    }

    String action = null;
    if(pushData != null) {
        action = pushData.optString("action", (String)null);
    }

    if(action != null) {
        Bundle notification = intent.getExtras();
        Intent broadcastIntent = new Intent();
        broadcastIntent.putExtras(notification);
        broadcastIntent.setAction(action);
        broadcastIntent.setPackage(context.getPackageName());
        context.sendBroadcast(broadcastIntent);
    }

    Notification notification1 = this.getNotification(context, intent);
    if(notification1 != null) {
        ParseNotificationManager.getInstance().showNotification(context, notification1);
    }

}

如果意图中没有额外的“警告”或“标题”，则不会发出通知。

因此，您根本不需要扩展任何类来进行静默推送更新......

如何覆盖ParsePushBroadcastReceiver的onPushReceive（）？

2 个答案: