Question

我有一个android应用程序。在那我有一个使用Firebase的推送通知。收到FirebaseMessagingService类的消息时，我需要在首页中显示弹出消息

FirebaseMessagingService

String actiondata = remoteMessage.getData().get("action");
if(actiondata.equals("booking")){
        Intent myIntent = new Intent("com.driver.approverequest");
        myIntent.putExtra("contentdata",contentdata);
        myIntent.putExtra("requestid",productId);
        this.sendBroadcast(myIntent);
        Log.d(TAG, "Message data : athira send " + message);
    }

家庭活动

 public BroadcastReceiver myReceiver = new BroadcastReceiver() {
    @Override
    public void onReceive(Context context, Intent intent) {
        String action = intent.getStringExtra("action");
        try {
            JSONObject mJsonObject = new JSONObject(intent.getExtras().getString("contentdata"));
            Log.d("hi...........",intent.getExtras().getString("contentdata"));
            Toast.makeText(getApplicationContext(),"hi",Toast.LENGTH_LONG).show();
            StartTrackingDriver();
        } catch (Exception e){

        }
    }
};

@Override
protected void onResume() {
    super.onResume();
    registerReceiver(myReceiver, new IntentFilter("com.driver.approverequest"));
}

@Override
protected void onPause() {
    super.onPause();
    try {
        unregisterReceiver(myReceiver);
    }catch (Exception e){}
}

Answer 1

在OnCreate()中注册广播

registerReceiver(myReceiver, new IntentFilter("com.driver.approverequest"));

在onDestroy()上取消注册

unregisterReceiver(myReceiver);

每当活动进入后台时，

在onPause中取消注册将取消广播的注册。

AND

您从意图中获得了字符串“ action”，但从未将值设置为intent。

String action = intent.getStringExtra("action");

因此，请先将字符串设置为intent。

myIntent.putExtra("action",actiondata);

编辑尝试在LocalBroadcastManager.getInstance(mContext)，register和unregister

之前使用sendbroadcast()

LocalBroadcastManager.getInstance(context).registerReceiver(myReceiver, new IntentFilter("com.driver.approverequest"));

LocalBroadcastManager.getInstance(context).unregisterReceiver(myReceiver);

LocalBroadcastManager.getInstance(context).sendBroadcast(myIntent);

Answer 2

@Arya Fcm致力于以下概念。

1个应用位于最前面 2个应用程序处于后台

1当应用程序位于最前面时，无论从服务器接收到的json有效负载如何，每次都会调用onReceive方法

2当您的应用程序在后台运行时，仅当您的json不包含“通知”键时，才会调用onReceive方法。如果您的有效负载包含通知密钥，则firebase sdk将管理所有内容。

Answer 3

String actiondata = remoteMessage.getData().get("action");
 if(actiondata.equals("booking")){
     Intent myIntent = new Intent("com.driver.approverequest");
     myIntent.putExtra("contentdata",contentdata);
     myIntent.putExtra("requestid",productId);
     LocalBroadcastManager.getInstance(this).sendBroadcast(myIntent);
     Log.d(TAG, "Message data : athira send " + message);
 }

//In your HomeActivity
@Override
 protected void onCreate(Bundle savedInstanceState) {
     super.onCreate(savedInstanceState);
  BroadcastReceiver mBroadcastReceiver = new BroadcastReceiver(){
  @Override
     public void onReceive(Context context, Intent intent) {
         String data = intent.getStringExtra("contentdata");
         String id = intent.getStringExtra("requestid");

 //Write your pop up message logic here
  }
     }

 @Override
 protected void onResume() {
 LocalBroadcastManager.getInstance(this).registerReceiver(mBroadcastReceiver,new IntentFilter("com.driver.approverequest"));
        super.onResume();
 }

 @Override
 protected void onPause() {  
  LocalBroadcastManager.getInstance(this).unregisterReceiver(mBroadcastReceiver);
     super.onPause();
 }

在FirebaseMessagingService中接收消息时弹出消息

3 个答案: