Deserialize XML Để đối tượng sử dụng Dynamic

Question

Có thể Deserialize XML chưa biết đối tượng như dưới đây không?

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

Bạn có thể thử điều này.

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 || node.HasAttributes ? (object)new DynamicXml(node) : node.Value; 
      return true; 
     } 

     return true; 
    } 
}