如何在派生自ObservableCollection的運行時中創建類的實例 <T> ？

Question

如何在派生自ObservableCollection的運行時中創建類的實例

視圖模型和模型如下：-C＃編碼

public class Mobile
{
    ObservableCollection<MobileModelInfo> SourceCollection = new ObservableCollection<MobileModelInfo>();

    private void CreateObject(ObservableCollection<MobileModelInfo> Source)
    {
        /// Create an Object for MobileModelInfo Class in Runtime and add the Values
    }

    private ObservableCollection<MobileModelInfo> CostructMobileModel()
    {
        SourceCollection.Add(new MobileModelInfo { Name = "iPhone 4", Catagory = "Smart Phone", Year = "2011" });
        SourceCollection.Add(new MobileModelInfo { Name = "S6", Catagory = "Ultra Smart Phone", Year = "2015" });

        CreateObject(SourceCollection);

        return SourceCollection;
    }

}

public class MobileModelInfo
{
    public string Name { get; set; }
    public string Catagory { get; set; }
    public string Year { get; set; }
}

Answer 1

從給出的示例中，您不需要ObservableCollection類型的后備變量； 只要繼承自它，您的Mobile類就成為MobileModelInfo的可觀察集合。 注意：使用以下設計模式將其綁定起來要容易得多。

public class Mobile : ObservableCollection<MobileModelInfo>
{
    public Mobile()
    {
        Add(new MobileModelInfo { Name = "foo", Category = "boo", Year = 1988 } );
    }

    public Mobile GetList()
    {
        return this;
    }
}

Answer 2

我找到了這個問題的解決方案。 以下是C＃函數，它在運行時從ObservableCollection派生一個類的實例

private void CreateObject(ObservableCollection<MobileModelInfo> Source)
{
    var gType = Source.GetType();
    string collectionFullName = gType.FullName;
    Type[] genericTypes = gType.GetGenericArguments();
    string className = genericTypes[0].Name;
    string classFullName = genericTypes[0].FullName;

    // Get the type contained in the name string
    Type type = Type.GetType(classFullName, true);

    // create an instance of that type
    object instance = Activator.CreateInstance(type);

    // List of Propery for the above created instance of a dynamic class
    List<PropertyInfo> oProperty = instance.GetType().GetProperties().ToList();
}