Question

Android的新手，试图弄清楚服务。我正在尝试将服务绑定到一个活动，我正在遵循文档中的示例，但我在下面标记的行（appService.playSong（title））上不断收到NullPointerException。在调试器中检查它会发现appService确实为null。

public class Song extends Activity implements OnClickListener,Runnable {
protected static int currentPosition;
private ProgressBar progress;
private TextView songTitle;
private MPService appService;

private ServiceConnection onService = new ServiceConnection() {
    public void onServiceConnected(ComponentName className,
            IBinder rawBinder) {
        appService = ((MPService.LocalBinder)rawBinder).getService();
    }

    public void onServiceDisconnected(ComponentName classname) {
        appService = null;
    }
};


public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.song);

    Intent bindIntent = new Intent(Song.this,MPService.class);
    bindService(bindIntent,onService,
            Context.BIND_AUTO_CREATE);

    Bundle b = getIntent().getBundleExtra("songdata");
    String title = b.getString("song title");

    // ... 

    appService.playSong(title); // nullpointerexception

    // ...

}

以下是该服务的相关部分：

package org.example.music;

// imports

public class MPService extends Service {
private MediaPlayer mp;
public static int currentPosition = 0;
public List<String> songs = new ArrayList<String>();
public static String songTitle;
private static final String MEDIA_PATH = new String("/mnt/sdcard/");

@Override
public void onCreate() {
    super.onCreate();

    mp = new MediaPlayer();
    songs = Music.songs;
}

@Override
public int onStartCommand(Intent intent, int flags, int startId) {
    return Service.START_STICKY;
}

public class LocalBinder extends Binder {
    MPService getService() {
        return MPService.this;
    }
}

private final IBinder binder = new LocalBinder();

@Override
public IBinder onBind(Intent intent) {
    return binder;
}

public void playSong(String songPath) {
try {
    mp.reset();
    mp.setDataSource(songPath);
    mp.prepare();
    mp.start();

    mp.setOnCompletionListener(new OnCompletionListener() {
        public void onCompletion(MediaPlayer arg0) {
            nextSong();
        }
    });

    songTitle = songPath.substring(12,songPath.length()-4);

} catch (IOException e) {
    Log.v(getString(R.string.app_name),e.getMessage());
}
}

public void nextSong() {
if (++currentPosition >= songs.size()) {
    currentPosition = 0;
}
String song = MEDIA_PATH+songs.get(currentPosition);
playSong(song);
} 

public void prevSong() {
if (--currentPosition<0) {
    currentPosition=songs.size()-1;
}
String song = Music.MEDIA_PATH+songs.get(currentPosition);
playSong(song);
}

public int getSongPosition() {
return mp.getCurrentPosition();
}

public MediaPlayer getMP() {
return mp;
}
}

我在AndroidManifest.xml中注册了该服务，并设置了android：enabled =“true”。你在这看到任何明显的错误吗？

Answer 1

您可以制作本地和远程两种绑定。本地仅供您的应用程序和远程使用，以供任何实现特定接口的应用程序使用。你应该从本地绑定开始。

Local binding tutorial。
Remote binding tutorial

我没有绑定的解决方案：

public class MyActivity extends Activity{

  @Override
  public void onCreate(Bundle savedInstanceState){
   ...
   Intent it = new Intent(MyService.ACTIVITY_START_APP);
   it.setClass(getApplicationContext(), MyService.class);
   startService(it);
}

  ...

@Override
    protected void onResume() {
        super.onResume();
        registerBroadcastReceiver();
    }

@Override
    protected void onPause() {
        super.onPause();
        this.unregisterReceiver(this.receiver);
    }

  ...

private BroadcastReceiver receiver = new BroadcastReceiver(){

        @Override
        public void onReceive(Context context, Intent intent) {
            if (intent.getAction().equals(MyService.BROADCAST_INIT)) {
                //do your stuff here after init
            }
        }
    };

private void registerBroadcastReceiver(){
         IntentFilter filter = new IntentFilter();
         filter.addAction(HMyService.BROADCAST_INIT);
         this.registerReceiver(receiver, filter);
    }
}

您的服务：

public class MyService extends Service{

public static final String BROADCAST_INITIAL_DATA = "org.myapp.BROADCAST_INIT";
public static final String ACTIVITY_START_APP = "org.myapp.ACTIVITY_START_APP";

  @Override
  public int onStartCommand (Intent intent, int flags, int startId){
    super.onStartCommand(intent, flags, startId);
    if(intent.getAction().equals(ACTIVITY_START_APP)){
      //do your initialization
      //inform the client/GUI
      Intent i = new Intent();
      i.setAction(BROADCAST_INIT);
      sendBroadcast(i);
    }else{
      //some other stuff like handle buttons
    }              
  }
}

祝你好运。

Answer 2

您假设bindService（）将同步连接到服务，但只有在onCreate（）finshed之后才能使用该连接。

框架在UI线程上运行onCreate（），而bindService（）只是记下以便稍后连接到服务。连接到服务将始终在UI线程上完成，因此这只能在执行onCreate之后进行。在onCreate（）之后，您甚至无法依赖正在建立的连接。它会发生在那之后的某个时间:)。此外，框架可能会断开服务的意愿，尽管它应该只在低内存条件下发生。

因此，将使用appService的代码从onCreate（）移动到onServiceConnected（），它就可以工作了。

Answer 3

从快速浏览一下，看起来您在绑定完成之前尝试访问您的服务。在尝试调用服务上的任何方法之前，您必须确保onServiceConnected已被解雇。

示例：

Intent bindIntent = new Intent(Song.this,MPService.class);
bindService(bindIntent,onService, Context.BIND_AUTO_CREATE);

//Wait until service has bound
while(appService == null){
     Thread.sleep(100);
}
appService.playSong(title);

此示例不是最佳示例，但它表明您必须等到绑定完成后再尝试访问该服务。

Android无法绑定到服务

3 个答案: