XML Serialization of List<T> - XML Root

Question

First question on Stackoverflow (.Net 2.0):

So I am trying to return an XML of a List with the following:

public XmlDocument GetEntityXml()
    {        
        StringWriter stringWriter = new StringWriter();
        XmlDocument xmlDoc = new XmlDocument();            

        XmlTextWriter xmlWriter = new XmlTextWriter(stringWriter);

        XmlSerializer serializer = new XmlSerializer(typeof(List<T>));

        List<T> parameters = GetAll();

        serializer.Serialize(xmlWriter, parameters);

        string xmlResult = stringWriter.ToString();

        xmlDoc.LoadXml(xmlResult);

        return xmlDoc;
    }

Now this will be used for multiple Entities I have already defined.

Say I would like to get an XML of List<Cat>

The XML would be something like:

<ArrayOfCat>
  <Cat>
    <Name>Tom</Name>
    <Age>2</Age>
  </Cat>
  <Cat>
    <Name>Bob</Name>
    <Age>3</Age>
  </Cat>
</ArrayOfCat>

Is there a way for me to get the same Root all the time when getting these Entities?

Example:

<Entity>
  <Cat>
    <Name>Tom</Name>
    <Age>2</Age>
  </Cat>
  <Cat>
    <Name>Bob</Name>
    <Age>3</Age>
  </Cat>
</Entity>

Also note that I do not intend to Deserialize the XML back to List<Cat>

Answer 1

If I understand correctly, you want the root of the document to always be the same, whatever the type of element in the collection ? In that case you can use XmlAttributeOverrides :

       XmlAttributeOverrides overrides = new XmlAttributeOverrides();
       XmlAttributes attr = new XmlAttributes();
       attr.XmlRoot = new XmlRootAttribute("TheRootElementName");
       overrides.Add(typeof(List<T>), attr);
       XmlSerializer serializer = new XmlSerializer(typeof(List<T>), overrides);
       List<T> parameters = GetAll();
       serializer.Serialize(xmlWriter, parameters);

Answer 2

A better way to the same thing:

public XmlDocument GetEntityXml<T>()
{
    XmlAttributeOverrides overrides = new XmlAttributeOverrides();
    XmlAttributes attr = new XmlAttributes();
    attr.XmlRoot = new XmlRootAttribute("TheRootElementName");
    overrides.Add(typeof(List<T>), attr);

    XmlDocument xmlDoc = new XmlDocument();
    XPathNavigator nav = xmlDoc.CreateNavigator();
    using (XmlWriter writer = nav.AppendChild())
    {
        XmlSerializer ser = new XmlSerializer(typeof(List<T>), overrides);
        List<T> parameters = GetAll<T>();
        ser.Serialize(writer, parameters);
    }
    return xmlDoc;
}

Answer 3

There is a much easy way:

public XmlDocument GetEntityXml<T>()
        {
            XmlDocument xmlDoc = new XmlDocument();
            XPathNavigator nav = xmlDoc.CreateNavigator();
            using (XmlWriter writer = nav.AppendChild())
            {
                XmlSerializer ser = new XmlSerializer(typeof(List<T>), new XmlRootAttribute("TheRootElementName"));
                ser.Serialize(writer, parameters);
            }
            return xmlDoc;
        }

Answer 4

so simple....

public static XElement ToXML(this IList lstToConvert, Func filter, string rootName) { var lstConvert = (filter == null) ? lstToConvert : lstToConvert.Where(filter); return new XElement(rootName,

            (from node in lstConvert
             select new XElement(typeof(T).ToString(),

                     from subnode in node.GetType().GetProperties()
                     select new XElement(subnode.Name, subnode.GetValue(node, null)))));

    }