Question

Is it possible to sort the elements and not only the nodes using XPathExpression.AddSort?

If I change the example code on the MSDN documentation for XPathExpression.AddSort a little so I request an element and not the whole node the sort order does not work.

var doc = new XPathDocument("contosoBooks.xml");
var nav = doc.CreateNavigator();

var expr = nav.Compile("/bookstore/book/title");

expr.AddSort("title", XmlSortOrder.Descending,
               XmlCaseOrder.None, "", XmlDataType.Number);

var iterator = nav.Select(expr);
while (iterator.MoveNext())
{
    Console.WriteLine(iterator.Current);
}

I would expect this output:

The Gorgias
The Confidence Man
The Autobiography of Benjamin Franklin

But the output is this:

The Autobiography of Benjamin Franklin
The Confidence Man
The Gorgias

Here is the xml file for your reference

<?xml version="1.0" encoding="utf-8" ?>
<bookstore>
  <book genre="autobiography" publicationdate="1981-03-22" ISBN="1-861003-11-0">
    <title>The Autobiography of Benjamin Franklin</title>
    <author>
      <first-name>Benjamin</first-name>
      <last-name>Franklin</last-name>
    </author>
    <price>8.99</price>
  </book>
  <book genre="novel" publicationdate="1967-11-17" ISBN="0-201-63361-2">
    <title>The Confidence Man</title>
    <author>
      <first-name>Herman</first-name>
      <last-name>Melville</last-name>
    </author>
    <price>11.99</price>
  </book>
  <book genre="philosophy" publicationdate="1991-02-15" ISBN="1-861001-57-6">
    <title>The Gorgias</title>
    <author>
      <name>Plato</name>
    </author>
    <price>9.99</price>
  </book>
</bookstore>

Answer 1

It looks like a duplicate question – see XPathExpression AddSort fails

So in your case code has to be like:

var expr = nav.Compile("/bookstore/book");

expr.AddSort("title", XmlSortOrder.Descending,
           XmlCaseOrder.None, "", XmlDataType.Number);

or

var expr = nav.Compile("/bookstore/book/title");

expr.AddSort(".", XmlSortOrder.Descending,
           XmlCaseOrder.None, "", XmlDataType.Number);