16

我创建了一个扩展 IntentService 的类,我想从一个不是 Activity 的类启动服务,因此,我无权访问 Context 对象。我在文档或网络中找不到这样的示例。是否可以 ?

4

2 回答 2

20

您需要将当前 Activity 上下文传递给非 Activity 类,以便从非 Activity 类启动服务:

public class NonActivity {
  public Context context;

  public NonActivity(Context context){
    this.context=context;
  }

  public void startServicefromNonActivity(){
     Intent intent=new Intent(context,yourIntentService.class);
     context.startService(intent);
  }
}

并将当前上下文传递为:

public class AppActivity extends Activity {
    NonActivity nonactiityobj;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
                 nonactiityobj=new NonActivity(CuttentActivity.this);
                 //start service here
                 nonactiityobj.startServicefromNonActivity();
    }
}
于 2012-12-14T16:03:45.137 回答
3

使用此代码启动和停止服务

public class MyService {

Context context ;

public MyService(Context cont) {
    this.context = context ;
}

public void StartMyService()
{
    Intent i = new Intent(context,YourService.class);
    context.startService(i);
}
public void StopMyService()
{
    Intent i = new Intent(context,YourService.class);
    context.stopService(i);
}
}

这只是创建这个类的对象

  MyService mySevice ;
  @Override
  public void onCreate(Bundle savedInstanceState) {
  super.onCreate(savedInstanceState);
  setContentView(R.layout.activity_main);
  myService = new MyService(this);
  //For Startting Service
  myService.StartMyService();

  //For Stopping Service
  myService.StopMyService();
}
于 2012-12-14T17:41:00.040 回答