动service
Intent intent = new Intent("com.Sevices.mqtt.ReceivingMessageService");
intent.putExtra("message", s);
startService(intent);
注册
<service android:name="com.Sevices.mqtt.ReceivingMessageService"/>
service的onStartCommand方法
public int onStartCommand(Intent intent, int flags, int startId) {
// TODO Auto-generated method stub
Log.i("serv1", "onStartCommand");
String s = intent.getStringExtra("message");
if (s != null && s.length() > 0)
{
Log.i("serv1", "message:"+s);
}
return super.onStartCommand(intent, flags, startId);
}
2013-11-07 14:43
提问者采纳
你只是简单的注册了<service android:name="com.Sevices.mqtt.ReceivingMessageService"/>
并没有给这个service起个名字，所以Intent intent = new Intent("com.Sevices.mqtt.ReceivingMessageService");
这样调用是启动不了你的service的，请改为类似如下：
Intent intent = new Intent(MainActivity.this,com.Sevices.mqtt.ReceivingMessageService.class);
或者在manifest中加个别名：
<service android:name="com.Sevices.mqtt.ReceivingMessageService">
<intent-filter>
<action android:name="com.Sevices.mqtt.ReceivingMessageService" />
</intent-filter>
</service>
提问者评价
-
谢谢!