java - 如何使用resultreceiver发送数据，而类是从线程扩展的

Question

我有一个服务类，我通过它向我的活动发送一些数据

public class baseApi extends Service {

@Override
      public int onStartCommand(Intent intent, int flags, int startId) {
  //get the receiver object
        resultReceiver = intent.getParcelableExtra("receiver");
}


  private BroadcastReceiver datachanged_receiver = new BroadcastReceiver() {

        @Override
        public void onReceive(Context context, Intent intent) {
                //HERE I send some data like this
                Bundle bundle = new Bundle(); 
                bundle.putString("data", "My data goes here");
                resultReceiver.send(100, bundle);
}};
}

在同一个类中，我有一个扩展“线程”的类，并且该类还打算使用 resultReceiver 向 UI 发送一些数据，它没有给出任何异常，但在 UI 端也没有收到数据，而 resultReceiver通过广播接收器发送数据效果很好。

public class baseApi extends Service {

@Override
      public int onStartCommand(Intent intent, int flags, int startId) {
  //get the receiver object
        resultReceiver = intent.getParcelableExtra("receiver");
}


  private BroadcastReceiver datachanged_receiver = new BroadcastReceiver() {

        @Override
        public void onReceive(Context context, Intent intent) {
                //HERE I send some data like this
                Bundle bundle = new Bundle(); 
                bundle.putString("data", "My data goes here");
                resultReceiver.send(100, bundle);
}};

class actionThread extends Thread{
    @Override
    public void run() {
        super.run();
        try{
                         //send this data to the service        
             Bundle bundle = new Bundle();
             bundle.putString("data","More data goes here" );
             resultReceiver.send(200, bundle);
        }
        catch(Exception ex)
        {

        }
   }

 }
}

Answer 1

在 android 中，您可以使用Handler与 UI 线程进行通信。只需使用Hanlder.PostorHandlerThread代替。