android - 我的onHandleIntent()没有被调用？

我有两个文件。

IntentServiceTest1Activity.java

 public class IntentServiceTest1Activity extends Activity {
     /** Called when the activity is first created. */

     public class ResponseReceiver extends BroadcastReceiver {

         public static final String ACTION_RESP = "com.mamlambo.intent.action.MESSAGE_PROCESSED";@Override
         public void onReceive(Context context, Intent intent) {
             // TODO Auto-generated method stub
             TextView result = (TextView) findViewById(R.id.txt_result);
             String text = intent.getStringExtra(SimpleIntentService.PARAM_OUT_MSG);
             result.setText(text);
         }
     }
     private ResponseReceiver receiver;

     @Override
     public void onCreate(Bundle savedInstanceState) {
         super.onCreate(savedInstanceState);
         setContentView(R.layout.main);

         IntentFilter filter = new IntentFilter(ResponseReceiver.ACTION_RESP);
         filter.addCategory(Intent.CATEGORY_DEFAULT);
         receiver = new ResponseReceiver();
         registerReceiver(receiver, filter);

         Button button = (Button) findViewById(R.id.button);
         button.setOnClickListener(new OnClickListener() {
             public void onClick(View v) {
                 EditText input = (EditText) findViewById(R.id.txt_input);
                 String strInputMsg = input.getText().toString();
                 Intent msgIntent = new Intent(IntentServiceTest1Activity.this, SimpleIntentService.class);
                 msgIntent.putExtra(SimpleIntentService.PARAM_IN_MSG, strInputMsg);
                 startService(msgIntent);
             }
         });
     }
 }

SimpleIntentService.java

 public class SimpleIntentService extends IntentService {

     public static final String PARAM_IN_MSG = "imsg";
     public static final String PARAM_OUT_MSG = "omsg";
     public SimpleIntentService()
     {
         super("SimpleIntentService");
     }
     @Override
     protected void onHandleIntent(Intent intent) {
         // TODO Auto-generated method stub
         String msg = intent.getStringExtra(PARAM_IN_MSG);
         SystemClock.sleep(30000); // 30 seconds
         String resultTxt = msg + " " + DateFormat.format("MM/dd/yy h:mmaa", System.currentTimeMillis());

         Intent broadcastIntent = new Intent();
         broadcastIntent.setAction(ResponseReceiver.ACTION_RESP);
         broadcastIntent.addCategory(Intent.CATEGORY_DEFAULT);
         broadcastIntent.putExtra(PARAM_OUT_MSG, resultTxt);
         sendBroadcast(broadcastIntent);
     }
 }

通过Log.d（），即使服务已启动，我也看不到onHandleIntent从未被调用。我的Log.d路径在startService（msgIntent）;之后；然后停在那里从onHandleIntent内部未显示Log.d。怎么了？我该怎么做才能使其正常工作？

谢谢！

最佳答案

我通过添加onStartCommand解决了我的问题，但没有重写它。

 public int onStartCommand(Intent intent, int flags, int startId) //sometimes overriding onStartCommand will not call onHandleIntent
{

    Logger_.logtoFile(tag, "here..!", 1);
    return super.onStartCommand(intent,flags,startId);
}

关于android - 我的onHandleIntent()没有被调用？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/11202101/