我已经定义了一个名为LeaderboardScore的ParseObject子类,它从我的云代码函数返回为IDictionary<string, object>
。
我希望我可以像下面的例子中那样做,但演员表失败了:(
投射失败的尝试:
ParseCloud.CallFunctionAsync<IDictionary<string, object>>("getScore", parameters).ContinueWith(t =>
{
LeaderboardScore score = t.result as LeaderboardScore;
Debug.Log(score.get<string>("facebookId"));
}
LeaderboardScore定义:
[ParseClassName("LeaderboardScore")]
public class LeaderboardScore : ParseObject
{
[ParseFieldName("facebookId")]
public string FacebookId
{
get { return GetProperty<string>("FacebookId"); }
set { SetProperty<string>(value, "FacebookId"); }
}
[ParseFieldName("score")]
public int Score
{
get { return GetProperty<int>("Score"); }
set { SetProperty<int>(value, "Score"); }
}
}
请注意,t.Result确实有正确的信息,这意味着我可以通过调用t.Result["facebookId"] as string
之类的东西来访问它,但是能够传递LeaderboardScore对象而不是{{1}会更好。 }}
如果有人能对这个问题有所了解,我将不胜感激! :)
答案 0 :(得分:1)
您可以通过以下方式将所有字典强制转换为对象(带属性):
public static T ToObject<T>(this IDictionary<string, object> source)
where T : class, new()
{
T someObject = new T();
Type someObjectType = someObject.GetType();
foreach (KeyValuePair<string, object> item in source)
{
someObjectType.GetProperty(item.Key).SetValue(someObject, item.Value, null);
}
return someObject;
}
答案 1 :(得分:0)
执行此操作的正确方法是为您感兴趣的每个ParseObject创建native subclasses,然后让您的云函数返回该类型或该类型的列表,如下所示:
ParseCloud.CallFunctionAsync<LeaderboardScore>("getScore", parameters).ContinueWith(t =>
{
LeaderboardScore score = t.Result;
}
Parse负责转换,因此您不必担心。要返回列表,只需使用IList<LeaderboardScore>