将XML反序列化为使用dynamic的对象

是否有可能反序列化未知的XML像下面的对象？

var xml = @"<Students><Student><Name>Arul</Name><Mark>90</Mark></Student></Students>"; var serializer = new XmlSerializer(typeof(DynamicObject)); dynamic students = serializer.Deserialize(new XmlTextReader(new StringReader(xml))); 

Answer 1

你可能想试试这个。

 string xml = @"<Students> <Student ID=""100""> <Name>Arul</Name> <Mark>90</Mark> </Student> <Student> <Name>Arul2</Name> <Mark>80</Mark> </Student> </Students>"; dynamic students = DynamicXml.Parse(xml); var id = students.Student[0].ID; var name1 = students.Student[1].Name; foreach(var std in students.Student) { Console.WriteLine(std.Mark); } 

---

 public class DynamicXml : DynamicObject { XElement _root; private DynamicXml(XElement root) { _root = root; } public static DynamicXml Parse(string xmlString) { return new DynamicXml(XDocument.Parse(xmlString).Root); } public static DynamicXml Load(string filename) { return new DynamicXml(XDocument.Load(filename).Root); } public override bool TryGetMember(GetMemberBinder binder, out object result) { result = null; var att = _root.Attribute(binder.Name); if (att != null) { result = att.Value; return true; } var nodes = _root.Elements(binder.Name); if (nodes.Count() > 1) { result = nodes.Select(n => n.HasElements ? (object)new DynamicXml(n) : n.Value).ToList(); return true; } var node = _root.Element(binder.Name); if (node != null) { result = node.HasElements ? (object)new DynamicXml(node) : node.Value; return true; } return true; } }