我们可以使这个方法通用吗？

Question

我在sample from Xamarin 中看到了这种方法，使用 JSON 访问 REST 服务器：

List<Country> countries = new List<Country>();
    public Task<List<Country>> GetCountries()
    {
        return Task.Factory.StartNew (() => {
            try {

                if(countries.Count > 0)
                    return countries;

                var request = CreateRequest ("Countries");
                string response = ReadResponseText (request);
                countries = Newtonsoft.Json.JsonConvert.DeserializeObject<List<Country>> (response);
                return countries;
            } catch (Exception ex) {
                Console.WriteLine (ex);
                return new List<Country> ();
            }
        });
    }

其中“CreateRequest”和“ReadResponseText”是与 REST 服务器交互的方法，基本上接收国家列表以反序列化并在列表中返回。 所以现在，我正在尝试使这个方法成为通用方法，以便接收类型并返回指定类型的对象的通用列表，如下所示：

public static Task<List<Object>> getListOfAnyObject(string requested_object, Type type) 
    {
        return Task.Factory.StartNew (() => {
            try {
                var request = CreateRequest (requested_object);
                string response = ReadResponseText (request);
                List<Object> objects = // create a generic list based on the specified type
                objects = Newtonsoft.Json.JsonConvert.DeserializeObject<List<Object>> (response); // not sure how to handle this line
                return objects;
            } catch (Exception ex) {
                Console.WriteLine (ex);
                return ex.Message;
            }
        });
    }

所以我的问题是，我怎样才能创建上面的方法以便越来越少地使用它（将列表转换为我想要的类型）？

List<Country> countries = (List<Country>)(List<?>) getListOfAnyObject("countries",Country.type);

非常感谢！

Answer 1

试试这样的..

public static Task<List<T>> getListOfAnyObject<T>(string requested_object) 
{
   return Task.Factory.StartNew (() => {
       try {
           var request = CreateRequest (requested_object);
           string response = ReadResponseText (request);
           return Newtonsoft.Json.JsonConvert.DeserializeObject<List<T>> (response); // not sure how to handle this line
       } catch (Exception ex) {
           Console.WriteLine (ex);
           return ex.Message;
       }
   });
}

这样称呼..

List<Country> countries = getListOfAnyObject<Country>("countries");