Question

I have a class PersonList

[XmlRoot("Persons")]
PersonList : List<Human>

when I serialize this to XML, by default it will produce something like this:

<Persons>
  <Human>...</Human>
  <Human>...</Human>
</Persons>

My question is what needs to be done in order to change element Human to Person in the output? so the output would be :

<Persons>
  <Person>...</Person>
  <Person>...</Person>
</Persons>

and, how to deserialize the above XML to the PersonList class object?

Per Nick’s advice, Here is my testing code:

[XmlRoot("Persons")]
public class Persons : List<Human>
{

}

[XmlRoot("Person")]
public class Human
{
    public Human()
    {
    }

    public Human(string name)
    {
        Name = name;
    }

    [XmlElement("Name")]
    public string Name { get; set; }

}

void TestXmlSerialize()
{
    Persons personList = new Persons();
    personList.Add(new Human("John"));
    personList.Add(new Human("Peter"));

    try
    {
        using (StringWriter writer = new StringWriter())
        {
            XmlSerializer serializer = new XmlSerializer(typeof(Persons));
            XmlWriterSettings settings = new XmlWriterSettings();
            settings.OmitXmlDeclaration = true;

            XmlSerializerNamespaces namespaces = new XmlSerializerNamespaces();
            namespaces.Add(string.Empty, string.Empty);

            XmlWriter xmlWriter = XmlWriter.Create(writer, settings);
            serializer.Serialize(xmlWriter, personList, namespaces);

            Console.Out.WriteLine(writer.ToString());
        }
    }
    catch (Exception e)
    {
        Console.Out.WriteLine( e.ToString());
    }
}

The output of the testing code is:

<Persons>
  <Human>
    <Name>John</Name>
  </Human>
  <Human>
    <Name>Peter</Name>
  </Human>
</Persons>

As the output shows, the [XmlRoot("Person")] on Human does not change the tag to Person from Human.

Answer 1

I don’t think there is a way for you to control the name of the generated array elements.

If you can however wrap the Persons collection inside another class you will then have complete control over the generated output using XmlArrayAttribute and XmlArrayItemAttribute.

If you cannot create this new class you can resort to implementing IXmlSerializable, but this is much more complex.

An example for the first alternative follows:

[XmlRoot("Context")]
public class Context
{
    public Context() { this.Persons = new Persons(); }

    [XmlArray("Persons")]
    [XmlArrayItem("Person")]
    public Persons Persons { get; set; }
}

public class Persons : List<Human> { }

public class Human
{
    public Human() { }
    public Human(string name) { Name = name; }
    public string Name { get; set; }
}

class Program
{
    public static void Main(string[] args)
    {
        Context ctx = new Context();
        ctx.Persons.Add(new Human("john"));
        ctx.Persons.Add(new Human("jane"));

        var writer = new StringWriter();
        new XmlSerializer(typeof(Context)).Serialize(writer, ctx);

        Console.WriteLine(writer.ToString());
    }
}