using XmlSerializer with unknown root element

Question

I have some XML files that look like this:

File "B156.xml"

<B156>
  <Customer>
    <Name>The Barn</Name>
    <Phone>0427825166</Phone>
  </Customer>
  <Orders>
    <Item>
      <Description>Black toner</Description>
      <Amount>$59.00</Amount>
    </Item>
  <Orders>
</B156>

File "B172.xml"

<B172>
  <Customer>
    <Name>Pixie Inc</Name>
    <Phone>0426553190</Phone>
  </Customer>
  <Orders>
    <Item>
      <Description>Colour toner</Description>
      <Amount>$79.00</Amount>
    </Item>
  <Orders>
</B172>

How do I use XmlSerializer in this case, considering the root element is dynamic and we cannot use XmlRoot("RootName") to specify it?

@silkfire I use XmlSerializer because there are a lot of properties that I need to serialise into domain class, examples above are simplified for brevity purpose. — Kagawa
– Kagawa, Commented Aug 28, 2020 at 5:48
considering your examples, cant you get the root element name from the file name itself? — Gurhan Polat
– Gurhan Polat, Commented Aug 28, 2020 at 7:58
@GurhanPolat I have no interest in getting the root element here. My problem is that XmlSerializer throws error if I don't specify a valid root, which in my case I can't because they are dynamic. (B156, B172 and so on) — Kagawa
– Kagawa, Commented Aug 28, 2020 at 12:30

tontonsevilla · Accepted Answer · 2020-08-28 09:57:29Z

1

You can try this,

CLASS OBJECT

[XmlRoot("root")]
public class CustomerOrder
{
    public Customer Customer { get; set; }

    [XmlArray("Orders")]
    [XmlArrayItem("Item")]
    public List<Order> Orders { get; set; }
}

public class Customer
{
    public string Name { get; set; }
    public string Phone { get; set; }
}

public class Order
{
    public string Description { get; set; }
    public string Amount { get; set; }
}

CHANGEROOT METHOD

    private static XmlDocument changeRoot(XmlDocument xmlDocument)
    {
        var newXmlDocument = new XmlDocument();
        var newRoot = newXmlDocument.CreateElement("root");
        newXmlDocument.AppendChild(newRoot);
        newRoot.InnerXml = xmlDocument.DocumentElement.InnerXml;

        return newXmlDocument;
    }

SERIALIZATION

        var xmlDocument = new XmlDocument();
        xmlDocument.Load("XMLFile1.xml");
        xmlDocument = changeRoot(xmlDocument);

        XmlSerializer serializer = new XmlSerializer(typeof(CustomerOrder));
        CustomerOrder result;

        using (TextReader reader = new StringReader(xmlDocument.InnerXml))
        {
            result = (CustomerOrder)serializer.Deserialize(reader);
        }

answered Aug 28, 2020 at 9:57

tontonsevilla

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1 Comment

Kagawa Over a year ago

This works! Sorry I forgot to accept this as answer earlier.

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