Android：活动和服务通信

Question

我有服务，必须通过BradcastReceiver与活动（MyActivity）进行通信。这是广播接收器声明intosame活动，以便：

private class MessaggiReceiver extends BroadcastReceiver { 
     public static final String ACTION_RELOAD = "it.android.dev.thecode.ACTION_RELOAD"; 
     @Override 
     public void onReceive(Context arg0, Intent intent) { 
      if(intent.getAction().equals(ACTION_RELOAD)){ 
       leggi_act; //Method of MyActivity 
      } 
     } 

    } 

但如何必须申报入清单包含MyActivity内brodcast？我用这个代码it.android.dev.thecode.MyActivity.MessaggiReceiver

<receiver android:name="it.android.dev.thecode.MyActivity.MessaggiReceiver"> 
      <intent-filter> 
       <action android:name="it.android.dev.thecode.ACTION_RELOAD"></action> 
      </intent-filter> 
</receiver> 

但其实是错误，然后我用这个代码it.android.dev.thecode.myactivity $ MessaggiReceiver

<receiver android:name="it.android.dev.thecode.MyActivity$MessaggiReceiver"> 
       <intent-filter> 
        <action android:name="it.android.dev.thecode.ACTION_RELOAD"></action> 
       </intent-filter> 
    </receiver> 

谢谢

Answer 1

您不需要在清单中注册广播接收器，而是通过代码在您的活动的onResume中注册它。不要忘记在onPause注销以避免任何泄漏。

@Override 
public void onResume() { 
    registerReceiver(mMessageReceiver, 
     new IntentFilter("it.android.dev.thecode.ACTION_RELOAD")); 
} 

@Override 
public void onPause() { 
    unregisterReceiver(mMessageReceiver); 
} 

private BroadcastReceiver mMessageReceiver = new BroadcastReceiver() { 
    @Override 
    public void onReceive(Context context, Intent intent) { 
    Log.d("receiver", "Got broadcast: " + intent.getAction()); 
    } 
};