从未扩展Activity的类中启动服务

Question

我创建了一个扩展IntentService的类，并且我想从一个不是Activity类的类中启动服务，因此我没有访问Context对象。我在文档或网络中找不到此类示例。这是否可能？

- Alex Cartwright

Answer 1

若想从非 Activity 类启动服务，则需要将当前 Activity 上下文传递给非 Activity 类，方法如下：

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();
    }
}

- ρяσѕρєя K

Answer 2

使用此代码开始和停止服务

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);
}
}

这只是创建该类的对象

  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();
}

- Mohd Saleem