我有一个界面:
public interface ICrawlService<T> where T : SocialPostBase
{
Task<int> Crawl(int accountId, Disguise disguise, ISocialAccountRepository socialAccountRepository, ISocialRepository<T> socialRepository, ISocialCrawlJobRepository jobRepository, IInstrumentationRepository instrumentationRepository);
}
我的社交资源库是:
public interface ISocialRepository<T> where T : class
{
IEnumerable<SocialPostCollection<T>> List { get; }
Task Add(SocialPostCollection<T> entity, string type);
Task AddPosts(List<T> entity, string type);
void Delete(SocialPostCollection<T> entity);
void Update(SocialPostCollection<T> entity);
T Find(string profileName, MediaType type);
}
我正在寻找一种多态设计,因此可以将不同的类实例化为一个类型。像这样:
var socialRepo = new SocialRepository<Video>(configration.CosmosDBServiceEndpoint, configration.CosmosDBSecret, configration.CosmosDBDatabaseId);
var socialRepo2 = new SocialRepository<Post>(configration.CosmosDBServiceEndpoint, configration.CosmosDBSecret, configration.CosmosDBDatabaseId);
ICrawlService<SocialPostBase> crawlService;
crawlService = new CrawlYoutubeProfileService();
var id = await crawlService.Crawl(jobId, null, _socialAccountRepo, socialRepo, _socialCrawlJobRepo, instrumentationRepo);
crawlService = new CrawlAnotherProfileService();
var id2 = await crawlService.Crawl(jobId, null, _socialAccountRepo, socialRepo2, _socialCrawlJobRepo, instrumentationRepo);
但是它将不接受通用参数的基类,但出现以下错误。
无法隐式转换类型 'SocialCrawlServices.CrawlYoutubeProfileService'到 “ SocialCrawlServices.ICrawlService”。 存在显式转换(您是否缺少演员表?)
那么您如何进行通用的多态设计?那不可能吗?
答案 0 :(得分:1)
那不可能吗?
不,有可能。您只需在out
中的通用参数之前添加ICrawlService
。
public interface ICrawlService<out T> where T : SocialPostBase
答案 1 :(得分:0)
错误显示“存在显式转换(您是否缺少演员表?)”。答案是将实现类显式转换为接口:
crawlService = (ICrawlService<SocialPostBase>)new CrawlYoutubeProfileService();
crawlService = (ICrawlService<SocialPostBase>)new CrawlAnotherProfileService();
只要CrawlYoutubeProfileService
和CrawlAnotherProfileService
具有实现SocialPostBase
的类型参数,就可以使用它,如下所示:
class YoutubePost : SocialPostBase
{
}
class CrawlYoutubeProfileService : ICrawlService<YoutubePost>
{
}