Question

如何在SL 5中将我的XML（文本）转换为场景（对象）。

    public static Scene Xml_to_Object(String xml_content)
    {


        Scene scene = null;


        try
        {
            XmlSerializer reader = new XmlSerializer(typeof(Scene));

            scene = new Scene();
            scene = (Scene)reader.Deserialize( xml_content ); //This is wrong!

        }
        catch (Exception e)
        {

        }
        finally
        {
            file.Close();
            file.Dispose();
        }


        return scene;

    }

xml_content将在哪里

var xml_content = "<Scene>...xml stuff here .. </Scene>";

场景场景= Scene.Xml_to_Object（xml_content）;

Answer 1

public T Deserialize<T>(string xml)
{
   using( var stream = new MemoryStream(Encoding.Unicode.GetBytes(xml)) )
   {
       var serializer = new DataContractSerializer(typeof (T));
       T theObject = (T)serializer.ReadObject(stream);
       return theObject;
    }
}

可以在DataContractSerializer中找到System.Runtime.Serialization。像这样使用它：

 Scene scene = Deserialize<Scene>(xml_content);

Answer 2

首先，您的类应具有为xml序列化所需的属性，并在类的每个属性上指定元素名称。 e.g。

[Serializable]
[XmlRoot("Scene")]
[XmlType]
public class Scene
{
    [XmlElement("Name")]
    public string Name { get; set; }

    [XmlElement("Type")]
    public string Type{ get; set; }
}

现在，您可以将其序列化为

 XmlSerializer xmlReq = new XmlSerializer(typeof(Scene));
 string xml = "<Scene><Name>Mahesh</Name><Type>ABC</Type></Scene>";
 var stream = new MemoryStream(Encoding.Unicode.GetBytes(xml));
 var resposnseXml = (Scene)xmlReq.Deserialize(stream);

使用Silverlight将XML序列化为类

2 个答案: