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我正在创建一个音乐播放器应用程序，我在其中创建了一个Service用于播放曲目的应用程序。

用于控制我正在使用的播放器bindService()

下面是代码，我使用 ServiceController 类来绑定（inistService() 内部）和 unbind（releaseService() 内部）：

public class ServiceController {

    MusicServiceAidl aidlObject;
    CallMService serviceConnection = new CallMService();
    Context context;

    public ServiceController(Context c) {
        this.context = c;
    }

    class CallMService implements ServiceConnection {

        @Override
        public void onServiceConnected(ComponentName name, IBinder boundService) {
            aidlObject = MusicServiceAidl.Stub
                    .asInterface((IBinder) boundService);
            Toast.makeText(context, "Service Connected",
                    Toast.LENGTH_SHORT).show();
        }

        @Override
        public void onServiceDisconnected(ComponentName paramComponentName) {
            aidlObject = null;
            Toast.makeText(context, "Service Disconnected",
                    Toast.LENGTH_SHORT).show();
        }
    }

    public void initService() {
        try {

            Intent serviceIntent = new Intent();
            serviceIntent.setClassName("com.example.async",
                    com.example.async.PlayTrack.class.getName());
            boolean ret = context.bindService(serviceIntent, serviceConnection,
                    Context.BIND_AUTO_CREATE);
            Toast.makeText(context, "Service bound with " + ret,
                    Toast.LENGTH_SHORT).show();
        } catch (Exception e) {
            Toast.makeText(context,
                    "initService(): " + e.toString(), Toast.LENGTH_LONG).show();
        }
    }

    public void releaseService() {
        context.unbindService(serviceConnection);
        serviceConnection = null;
        Toast.makeText(context, "Service Released",
                Toast.LENGTH_SHORT).show();
    }

    public String getTrackName() throws RemoteException{
        return aidlObject.getTrackName();
    }

    public String getAlbumName() throws RemoteException{
        return aidlObject.getAlbumName();
    }

    public String getArtistName() throws RemoteException{
        return aidlObject.getArtistName();
    }

    public void playPreviousTrack() throws RemoteException{
        aidlObject.playPreviousTrack();
    }

    public void playNextTrack() throws RemoteException{
        aidlObject.playNextTrack();
    }

}

要调用这个绑定类，我正在使用：

ServiceController serviceController = new ServiceController(getApplicationContext());
serviceController.initService();
serviceController.releaseService();

问题是我试图从不同的类停止服务，即我想releaseService从不同的类调用。但很明显，它给出了IllegalArgumentException.

编辑： 当我运行以下代码时：

public void onBackPressed() {
    try {
        Intent intent = new Intent(this,
                Class.forName("com.example.async.PlayTrack"));
        stopService(intent);
        serviceController.releaseService();
    } catch (Exception e) {
        Log.e("Listing.java", e.toString());
        Toast.makeText(Listing.this,
                "Listing->onBackPressed: " + e.toString(), Toast.LENGTH_SHORT)
                .show();
    }
}

我得到以下异常

java.lang.IllegalArgumentException: Service not registered: com.example.async.classes.ServiceController$CallMService@40547298

我怎样才能做到这一点？

4

1 回答 1

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使用应用程序上下文而不是活动上下文进行绑定和取消绑定。您可以使用getApplicationContext().

于 2012-08-14T09:03:36.003 回答