deserialize xml attribute of xmlarray

Question

I know you can't add an attribute to an XmlArray which I think is really unhandy. I know I can make a separate class for Phone, however, Phone belongs in Access. I have about 30 more nodes under access. How can I deserialize attribute hasTextField?

    <Access>
       <Phone hasTextField="true">
          <Item description="Skype" />
          <Item description="IP Phone" />
       </Phone>
       <Computer>
          <Item description="PC" />
          <Item description="Laptop" />
       </Computer>
    </Access>


   [XmlRoot("Access")]

    public class Access
    {
       public Access(){}

       [XmlArray("Phone")]
       [XmlArrayItem("Item")]
       public AccessItem[] ItemList;

       [XmlArray("Computer")]
       [XmlArrayItem("Item")]
       public AccessItem[] ItemList;
    }

Is this read from an xml file? Have you considered using XmlReader? You can choose which properties are assigned that way. — lehn0058, Jul 26 '13 at 14:41
You may decorate both Phone and Computer classes with appropriate attribute (=do not use them in the list itself, moreover it doesn't even compile so...) — Adriano Repetti, Jul 26 '13 at 14:41
I believe this is a duplicate of this: http://stackoverflow.com/questions/1012360/how-to-add-attributes-for-c-sharp-xml-serialization — Kai Hartmann, Jul 26 '13 at 14:43
@Kai It's similar, however, the solution in that post is to create a new class, which I'd rather not do, because then I have to create a class for every node. — Robert Vogelezang, Jul 26 '13 at 15:01

score 4 · Accepted Answer · edited May 23 '17 at 12:29

You can replace the XmlArray with an XmlElement.

See: How to add an attribute to a collection marked with XmlArrayAttribute?

[XmlType("Access")]
public class Access
{
   [XmlElement("Phone")]
   public AccessItem Phone { get; set; }

   [XmlElement("Computer")]
   public AccessItem Computer { get; set; }
}

public class AccessItem
{
    public AccessItem()
    {
        Items = new List<Item>();
    }

    [XmlAttribute("hasTextField")]
    public bool HasTextField { get; set; }

    [XmlElement("Item")]
    public List<Item> Items { get; set; }
}

[XmlType("Item")]
public class Item
{
    [XmlAttribute("description")]
    public string Description { get; set; }
}

Code:

var data = @"<Access>
   <Phone hasTextField=""true"">
      <Item description=""Skype"" />
      <Item description=""IP Phone"" />
   </Phone>
   <Computer>
      <Item description=""PC"" />
      <Item description=""Laptop"" />
   </Computer>
</Access>";

var serializer = new XmlSerializer(typeof(Access));

Access access;

using(var stream = new StringReader(data))
using(var reader = XmlReader.Create(stream))
{
    access = (Access)serializer.Deserialize(reader);
}

Nice, just change computer to be computer, and phone to be phone :) — Sergey Berezovskiy, Jul 26 '13 at 15:13
Yeah I didn't notice that. I also changed the fields to properties. — Dustin Kingen, Jul 26 '13 at 15:15
Thanks! Using XmlElement works fabulous, I thought of it but just didn't know how to code it. — Robert Vogelezang, Jul 26 '13 at 15:35

deserialize xml attribute of xmlarray

1 Answers1