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Home/ Questions/Q 8541355
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T11:48:47+00:00 2026-06-11T11:48:47+00:00

I’m having trouble producing the following XML-structure using the XmlSerializer: <Root xmlns:xsi=http://www.w3.org/2001/XMLSchema-instance xmlns:xsd=http://www.w3.org/2001/XMLSchema> <Name>This

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I’m having trouble producing the following XML-structure using the XmlSerializer:

<Root xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Name>This is root.</Name>
  <OtherValue>Otha.</OtherValue>
  <Shapes Name="This attribute is ignored!">
    <Circle>
      <Name>This</Name>
      <Value>Is</Value>
      <Whatever>Circle</Whatever>
    </Circle>
    <Square>
      <Name>And</Name>
      <Value>this is</Value>
      <Something>Square</Something>
    </Square>
  </Shapes>
</Root>

The only problem is that the attributes of <Shapes> doesn’t get written.
The classes I’m using for the serialization are the following:

public class Root
{
    [XmlElement]
    public string Name { get; set; }

    [XmlElement]
    public string OtherValue { get; set; }

    [XmlArray("Shapes")]
    [XmlArrayItem("Circle", typeof(Circle))]
    [XmlArrayItem("Square", typeof(Square))]
    public ShapeList Shapes { get; set; }
}

public class ShapeList : List<Shape>
{
    // Attribute that is not in output
    [XmlAttribute]
    public string Name { get; set; }
}

public class Shape
{
    public string Name { get; set; }
    public string Value { get; set; }
}

public class Circle : Shape
{
   public string Whatever { get; set; }
}

public class Square : Shape
{
    public string Something { get; set; }
}

And a little main method to run the serialization:

public static void Main(String[] args)
{
    var extraTypes = new Type[] {
        typeof(Shape),
        typeof(Square),
        typeof(Circle)
    };

    var root = new Root();
    root.Name = "This is root.";
    root.OtherValue = "Otha.";

    root.Shapes = new ShapeList()
    {
        new Circle() { Name = "This", Value="Is", Whatever="Circle" },
        new Square() { Name = "And", Value="this is", Something="Square" }
    };
    root.Shapes.Name = "This is shapes.";

    using (var sw = new StreamWriter("data.xml"))
    {
        var serializer = new XmlSerializer(typeof(Root), extraTypes);
        serializer.Serialize(sw, root);
    }
}
  • Why am I not getting the Name attribute of the ShapeList?
  • If this cannot be accomplished using this method, is there any another simple way?
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T11:48:48+00:00Added an answer on June 11, 2026 at 11:48 am

    Attributes are not processed for the outer part of arrays; only leaf nodes are processed for that – collections are just: their contents. There is a way to do it, if you don’t mind making the model a bit more complex….

    public class Root
    {
        [XmlElement]
        public string Name { get; set; }
    
        [XmlElement]
        public string OtherValue { get; set; }
    
        [XmlElement("Shapes")]
        public ShapeContainer Shapes { get; set; }
    }
    
    public class ShapeContainer
    {
        [XmlAttribute]
        public string Name { get; set; }
    
        private readonly List<Shape> items = new List<Shape>();
        [XmlElement("Circle", typeof(Circle))]
        [XmlElement("Square", typeof(Square))]
        public List<Shape> Items { get { return items; } }
    }
    
    public class Shape
    {
        public string Name { get; set; }
        public string Value { get; set; }
    }
    
    public class Circle : Shape
    {
        public string Whatever { get; set; }
    }
    
    public class Square : Shape
    {
        public string Something { get; set; }
    }
    

    with usage changed as follows:

    root.Shapes = new ShapeContainer();
    root.Shapes.Items.Add(new Circle() { Name = "This", Value="Is", Whatever="Circle" });
    root.Shapes.Items.Add(new Square() { Name = "And", Value = "this is", Something = "Square" });
    root.Shapes.Name = "This is shapes.";
    
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