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,我无法访问Context对象。我在文档或网络上找不到这样的例子。有可能吗?
答案 0 :(得分:19)
您需要将当前的Activity上下文传递给非Activity类,以便从非活动类启动服务:
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();
}
}
答案 1 :(得分:2)
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();
}