Object position (line, column) in XML after deserialization .NET

Question

How can I get position in the original xml file of an xml tag after deserialization into a .NET object using XmlSerializer ?

Here is an example XML

Answer 1

Another, more simple approach: Let the deserializer do the work.

• Add LineInfo and LinePosition properties to all classes for which you would like to have position information:

  [XmlRoot("Root")]
  public class AddressDetails
  {
      [XmlAttribute]
      public int LineNumber { get; set; }
  
      [XmlAttribute]
      public int LinePosition { get; set; }
      ...
  }
  

  This of course can be done by subclassing.

• Load an XDocument with LoadOptions.SetLineInfo.

• Add LineInfo and LinePosition attributes to all elements:

  foreach (var element in xdoc.Descendants())
  {
       var li = (IXmlLineInfo) element;
       element.SetAttributeValue("LineNumber", li.LineNumber);
       element.SetAttributeValue("LinePosition", li.LinePosition);
   }
  

• Deserializing will populate LineInfo and LinePosition.

Cons:

• Line information only for elements that are deserialized as class, not for simple elements, not for attributes.
• Need to add attributes to all classes.