c# - 使用 LINQ 创建一个 List<T> 其中 T : someClass<U>

Question

这与我之前的一个问题有关 C# Generic List conversion to Class implementing List<T>

我有以下代码:

public abstract class DataField
{
    public string Name { get; set; }
}

public class DataField<T> : DataField
{
    public T Value { get; set; }
}

public static List<DataField> ConvertXML(XMLDocument data) {  
     result = (from d in XDocument.Parse(data.OuterXML).Root.Decendendants()
                      select new DataField<string>
                      {
                          Name = d.Name.ToString(),
                          Value = d.Value
                      }).Cast<DataField>().ToList();  
    return result;
}

这有效，但是我希望能够将 LINQ 查询的选择部分修改为如下所示:

select new DataField<[type defined in attribute of XML Element]>
{
  Name = d.Name.ToString(),
  Value = d.Value
}

这只是一个糟糕的方法吗？可能吗？有什么建议吗？

Answer 1

这是一个可行的解决方案:(您必须为 Type 属性指定完全限定的类型名称，否则您必须以某种方式配置映射...)

我使用了 dynamic 关键字，如果你没有 C# 4，你可以使用反射来设置值......

public static void Test()
{
    string xmlData = "<root><Name1 Type=\"System.String\">Value1</Name1><Name2 Type=\"System.Int32\">324</Name2></root>";

    List<DataField> dataFieldList = DataField.ConvertXML(xmlData);

    Debug.Assert(dataFieldList.Count == 2);
    Debug.Assert(dataFieldList[0].GetType() == typeof(DataField<string>));
    Debug.Assert(dataFieldList[1].GetType() == typeof(DataField<int>));
}

public abstract class DataField
{
    public string Name { get; set; }

    /// <summary>
    /// Instanciate a generic DataField<T> given an XElement
    /// </summary>
    public static DataField CreateDataField(XElement element)
    {
        //Determine the type of element we deal with
        string elementTypeName = element.Attribute("Type").Value;
        Type elementType = Type.GetType(elementTypeName);

        //Instanciate a new Generic element of type: DataField<T>
        dynamic dataField = Activator.CreateInstance(typeof(DataField<>).MakeGenericType(elementType));
        dataField.Name = element.Name.ToString();

        //Convert the inner value to the target element type
        dynamic value = Convert.ChangeType(element.Value, elementType);

        //Set the value into DataField
        dataField.Value = value;

        return dataField;
    }

    /// <summary>
    /// Take all the descendant of the root node and creates a DataField for each
    /// </summary>
    public static List<DataField> ConvertXML(string xmlData)
    {
        var result = (from d in XDocument.Parse(xmlData).Root.DescendantNodes().OfType<XElement>()
                      select CreateDataField(d)).ToList();

        return result;
    }
}

public class DataField<T> : DataField
{
    public T Value { get; set; }
}