XPath и атрибуты

[Luzifer]

Уважаемые гуру! Подскажите. Есть xml типа:

<root>
 <items>
  <item id="..." name="..."/>
  <item id="..." name="..."/>
  <item id="..." name="..."/>
 </items>
 ...
</root>

Надо ввести запрос XPath к примеру /root/items/* и должен быть оутпут List<string> типа:
list[0] = "<item id=\"...\" name=\"...\"/>"
list[1] = "<item id=\"...\" name=\"...\"/>"
list[2] = "<item id=\"...\" name=\"...\"/>"
...

Вот участок кода, который надо допилить:

public static List<String> ExecuteQuery(String pathFile, String query)
{
 XPathDocument xPathDocument = new XPathDocument(pathFile);
 XPathNavigator xPathNavigator = xPathDocument.CreateNavigator();
 XPathNodeIterator xPathNodeIterator = xPathNavigator.Select(query);
 
 List<String> tempList = new List<string>();
 while (xPathNodeIterator.MoveNext())
  tempList.Add(xPathNodeIterator.Current.Value);
 
 return tempList;
}

[Luzifer]

тему можно оффнуть. Решил так:

        public static List<String> ExecuteQuery(String pathFile, String query)
        {
            XPathDocument xPathDocument = new XPathDocument(pathFile);
            XPathNavigator xPathNavigator = xPathDocument.CreateNavigator();
            XPathExpression xPathExpression = xPathNavigator.Compile(query);
            XPathNodeIterator xPathNodeIterator = xPathNavigator.Select(xPathExpression);

            List<String> tempList = new List<string>();
            while (xPathNodeIterator.MoveNext())
                tempList.Add(xPathNodeIterator.Current.OuterXml);

            return tempList;
        }