c# - 重构这个 : Generic extension method

Question

我有一个扩展方法来检查对象的类型，然后填充它的成员属性

public static void LoadMeeting<T>(this T entity, IMeetingRepository meetingRepository) 
where T: MyEntity
            {
                var agenda = entity as Agenda;

                if (agenda != null)
                {
                    agenda.Meeting = meetingRepository.GetMeetingById(agenda.MeetingId);
                }

                var participant = entity as Participant;

                if (participant != null)
                {
                    participant.Meeting = meetingRepository.GetMeetingById(participant.MeetingId);
                }

            }

我可以进一步将其重构为这样的东西以使其更通用吗？

 public static void LoadMeeting<T>(this T entity, IMeetingRepository meetingRepository) where T : MyEntity
        {
            var obj = entity as Agenda || entity as Participant;

            if (obj != null)
            {
                obj.Meeting = meetingRepository.GetMeetingById(obj.MeetingId);
            }
        }
    }

PS:我不想把对象的属性Meeting放在基类(MyEntity)

Answer 1

就我个人而言，我只会使用重载:

public static void LoadMeeting(this Agenda agenda, IMeetingRepository meetingRepository) 
{
    if (agenda != null)
    {
         agenda.Meeting = meetingRepository.GetMeetingById(agenda.MeetingId);
    }
}
public static void LoadMeeting(this Participant participant, IMeetingRepository meetingRepository) 
{
    if (participant != null)
    {
        participant.Meeting = meetingRepository.GetMeetingById(participant.MeetingId);
    }
}

除非您有一些提供 Meeting 属性的共享协定(实现的基类或接口(interface))，否则您无法使用单个泛型方法来完成此操作。

另一种方法是制作一个共享界面，即:IMeeting，然后限制为:

public interface IMeeting
{
   public Meeting Meeting { get; set; }
   public int MeetingId { get; }
}

然后你可以这样写:

public static void LoadMeeting<T>(this T entity, IMeetingRepository meetingRepository) 

其中 T: IMeeting { 如果(实体!= null) { entity.Meeting = meetingRepository.GetMeetingById(entity.MeetingId); }

当您使用 EF 时，您可以在分部类中实现此接口(interface):

public partial class Agenda : MyEntity, IMeeting
{
}