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Home/ Questions/Q 7493841
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T17:12:16+00:00 2026-05-29T17:12:16+00:00

I want to deserialize a XML file in C# (.net 2.0). The structure of

  • 0

I want to deserialize a XML file in C# (.net 2.0).

The structure of the XML is like this:

<elements>
   <element>
     <id>
       123
     </id>
     <Files>
       <File id="887" description="Hello World!" type="PDF">
         FilenameHelloWorld.pdf
       </File>
     </Files>
   </element>
<elements>

When I try to deserialize this structure in C#, I get a problem with the Filename, the value is always NULL, even how I try to code my File class.

Please help me. 😉

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T17:12:17+00:00Added an answer on May 29, 2026 at 5:12 pm

    The following works fine for me:

    public class element
    {
        [XmlElement("id")]
        public int Id { get; set; }
    
        public File[] Files { get; set; }
    }
    
    public class File
    {
        [XmlAttribute("id")]
        public int Id { get; set; }
    
        [XmlAttribute("description")]
        public string Description { get; set; }
    
        [XmlAttribute("type")]
        public string Type { get; set; }
    
        [XmlText]
        public string FileName { get; set; }
    }
    
    class Program
    {
        static void Main()
        {
            using (var reader = XmlReader.Create("test.xml"))
            {
                var serializer = new XmlSerializer(typeof(element[]), new XmlRootAttribute("elements"));
                var elements = (element[])serializer.Deserialize(reader);
    
                foreach (var element in elements)
                {
                    Console.WriteLine("element.id = {0}", element.Id);
                    foreach (var file in element.Files)
                    {
                        Console.WriteLine(
                            "id = {0}, description = {1}, type = {2}, filename = {3}", 
                            file.Id,
                            file.Description,
                            file.Type,
                            file.FileName
                        );
                    }
                }
    
            }
        }
    }
    
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