Android服务不会立即启动

如何解决Android服务不会立即启动

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public class TestService extends Activity {
    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        final Intent service=new Intent(getApplicationContext(),MessageListener.class);

        Log.v(\"Test\",\"Going to start service\");    
        startService(service);
        Log.v(\"Test\",\"service started?\");

    }
}

public class MessageListener extends Service {

    @Override
    public int onStartCommand(Intent intent,int flags,int startId) {
        Log.v(\"Test\",\"Start Cmd\");
        intent.setAction(\"Started\");
        new Thread(new Runnable() {

            @Override
            public void run() {
                for(int i=100;i<200;i++){
                    Log.v(\"Test\",i+\"\");
                }

            }
        }).start();
        return START_STICKY;

    }

    @Override
    public void onCreate() {
        super.onCreate();
        Log.v(\"Test\",\"Create\");
    }

我希望它会打印：

Start Service
create
Start cmd
print 1->100
Service Started.

但是我越来越

Start Service
Service Started.
create 
Start cmd
prints 1->100

为什么？ 我发现问题是由于异步引起的。父方法完成后，将调用startService。 解决方案是：

public class TestService extends Activity {
   /** Called when the activity is first created. */
   @Override
   public void onCreate(Bundle savedInstanceState) {
       super.onCreate(savedInstanceState);
       setContentView(R.layout.main);
       Intent service=new Intent(getApplicationContext(),MessageListener.class);

               startService(service);

               mCheckerHandler.sendEmptyMessageDelayed(MSG_CHECK_SERVICE_RUNNING,100);
   }

   private static final int MSG_CHECK_SERVICE_RUNNING = 0x001122;

   private Handler mCheckerHandler = new Handler() {
           public void handleMessage(android.os.Message msg) {
                   if (msg.what == MSG_CHECK_SERVICE_RUNNING) {
                           if (checkServiceRunning()) {
                                   //Do something
                           } else {
                                   //Send another message to check in the next 100ms
                                   sendEmptyMessageDelayed(MSG_CHECK_SERVICE_RUNNING,100);
                           }
                   }
           };
   };
}

谢谢大家特别是对平先生：)     

解决方法

那是因为线程是在“伪”并行执行的，所以在计数器线程获得任何CPU时间来写入之前调用

Log.v(\"Test\",\"service started?\");

。 \“ Pseudo \”（并行），因为大多数电话没有超过1个CPU，因此它们无法并行计算，因此只能从一个线程切换到另一个线程。您可以在Wikipedia或您喜欢的任何其他来源上阅读有关线程的更多信息。