Question

情况：

活动绑定到已启动的前台服务。
服务将本地活页夹分发给活动。
activity通过getService（）调用获取对服务的引用。
活动希望使用消息与服务中运行的线程直接进行通信。它从活动中调用mService.getThreadHandler（）方法。

问题：

如何从当前运行的线程中获取Handler到活动活动中，以便我可以直接将消息发布到线程messagequeue？

我不知道在服务中使用什么信使，我想直接与活动方面的服务中的线程进行通信。

编辑：活动通过调用以下内容获取服务中线程本身的处理程序：

活动代码：

Handler mServiceThreadHandler;
ServiceConnection mServiceConnection;

public void onStart() {
  if (!bindService(new Intent(this, MainService.class), mServiceConnection,    Context.BIND_AUTO_CREATE)) 
  {
    Log.e(TAG, "Client did not bind to Service!");
  } 
  super.onStart();
}

public class MyLocalServiceConnection implements ServiceConnection {
    @Override
    public void onServiceConnected(ComponentName name, IBinder service) {
        LocalBinder binder = (LocalBinder) service;
        mService = (MainService) binder.getService();
                    // the handler of the service is actually a handler of a thread
                    // within the service, and is set automatically within the binding
                    // activity when binding to the service. That way you have a direct
                    // "connection" with the message queue of the thread instead of
                    // a message queue in the service itself (main thread of service)
        mServiceThreadHandler = mService.getServiceHandler();

        if (mServiceThreadHandler == null) {
            Log.e(TAG, "Service handler is NULL");
        }
        mBoundedToService = true;
    }

    @Override
    public void onServiceDisconnected(ComponentName name) {
        mServiceThreadHandler = null;
        mBoundedToService = false;
    }
}

服务代码：

private HandlerThread mServiceThread = new MyServiceThread();

public Handler getServiceHandler() {
    return new Handler(mServiceThread.getLooper());
}

new Handler（mServiceThread.getLooper（））; 是否在mServiceThread中返回一个新的Handler或同一个Handler？

编辑2：使用接收邮件的serviceThread更新服务代码。

public class MyService extends Service {
  private MyHandlerThread serviceThread = new MyHandlerThread("serviceThread");

  serviceThread.start();

public Handler getServiceHandler() {
  // hand out the same handler of the service thread for using it in an activity!
  // serviceThread.getLooper() is the current looper of the thread
  // serviceThread is the 'this' which handles the messages (see MyHandlerThread)
  return new Handler(serviceThread.getLooper(), serviceThread); 
}

  // do stuff in Service
}

public class MyHandlerThread extends HandlerThread implements android.os.Handler.Callback {
public MyHandlerThread(String name) {
    super(name);
}
public MyHandlerThread(String name, int priority) {
    super(name, priority);
}
@Override
public boolean handleMessage(Message msg) {
    // TODO define your own message handling here.
    return false;
}
}

正确？

Answer 1

试试这个（我使用Activity来测试它你将使用Service）：

protected void onCreate(Bundle savedInstanceState) {
    ht = new HandlerThread("HT");
    ht.start();
    htHandler = new Handler(ht.getLooper(), htCallback);
    mainHandler = new Handler(mainCallback);
    for (int i = 0; i < 10; i++) {
        htHandler.sendMessageDelayed(htHandler.obtainMessage(99, i, 0), i * 3000);
    }
}

private HandlerThread ht;
private Handler htHandler;
private Handler mainHandler;

private Callback htCallback = new Callback() {
    @Override
    public boolean handleMessage(Message msg) {
        Log.d(TAG, "handleMessage **********************");
        Log.d(TAG, "handleMessage " + msg);
        Log.d(TAG, "handleMessage Thread: " + Thread.currentThread());
        if (msg.arg1 == 4) {
            Log.d(TAG, "handleMessage sending back to Main thread");
            mainHandler.sendEmptyMessage(101);
        }
        return false;
    }
}; 

private Callback mainCallback = new Callback() {
    @Override
    public boolean handleMessage(Message msg) {
        Log.d(TAG, "handleMessage ########################");
        Log.d(TAG, "handleMessage " + msg);
        Log.d(TAG, "handleMessage Thread: " + Thread.currentThread());
        Log.d(TAG, "handleMessage i'm quitting");
        ht.quit();
        return false;
    }
};

从服务到活动的线程获取处理程序

1 个答案: