Question

我是android的初学者。尝试从MainActivity启动服务但它失败了。这是调用方法按钮onClick listener

@Override
            public void onClick(View v)
            {
                Intent i = new Intent(getApplicationContext(), MyService.class);
                try
                {
                    startActivity(i);
                }
                catch(Exception e)
                {
                    Toast.makeText(getApplicationContext(), "Failed to start MyService", Toast.LENGTH_SHORT).show();
                }
            }

服务如下

public class MyService extends Service

{

@Override
public void onStart(Intent intent, int startId)
{
    // TODO: Implement this method
    super.onStart(intent, startId);

    Toast.makeText(getApplicationContext(), "onStart method in MyServive class", Toast.LENGTH_SHORT).show();
}

@Override
public IBinder onBind(Intent p1)
{
    // TODO: Implement this method
    Toast.makeText(getApplicationContext(), "onBind method in MyServive class", Toast.LENGTH_SHORT).show();

    return null;
}

@Override
public void onCreate()
{
    // TODO: Implement this method
    super.onCreate();

    Toast.makeText(getApplicationContext(), "onCreate method in MyServive class", Toast.LENGTH_SHORT).show();
}

@Override
public int onStartCommand(Intent intent, int flags, int startId)
{
    // TODO: Implement this method
    Toast.makeText(getApplicationContext(), "onStartCommand method in MyServive class", Toast.LENGTH_SHORT).show();

    return super.onStartCommand(intent, flags, startId);
}

@Override
public void onDestroy()
{
    // TODO: Implement this method
    super.onDestroy();

    Toast.makeText(getApplicationContext(), "onDestroy method in MyServive class", Toast.LENGTH_SHORT).show();
}

}

清单声明在

之下

<Service
        android:enabled="true"
        android:name=".MyService"
        android:label="MyService"
        >
    </Service>

可能会让它无法启动？没有为任何方法显示吐司

Answer 1

        @Override
        public void onClick(View v)
        {
            Intent i = new Intent(getApplicationContext(), MyService.class);
            try
            {
                startService(i);

            }
            catch(Exception e)
            {
                Toast.makeText(getApplicationContext(), "Failed to start MyService", Toast.LENGTH_SHORT).show();
            }
        }

服务无法在Android中启动

1 个答案: