XMLdocument Sort

Question

I've figured out how to append nodes to my rss document in the right structyre. I now need to sort it in the pubDate order and then output to screen. Looking at the examples online, I've found lots of XDocument and Linq stuff but nothing with XmlDocument. Scratching my head whether to scrap what code I have and work out how to do it in XDocument with advice from here or continue with XMLDocument and figure out a way to sort.

With XMLDocument I've got the code working exactly as I want, just need my feed to be sorted in pubDate order when it spits it out to the screen. So I think I will stick with this for the timebeing. I've found this article http://support.microsoft.com/kb/555060 and an xslt someone posted in Stack Overflow, but I dont know how to call the "XmlHelperFunctions" from my code. Is XSLT the easiest option I have, or is there something easier out there?

This is my code:

    XmlDocument xmlDoc = new XmlDocument();

    xmlDoc.LoadXml(rssFeed.ToString());

    XmlNodeList nl = xmlDoc.SelectNodes("/rss/channel/item");

    foreach (XmlNode xn in nl)
    {
        string title = xn["title"].InnerText;
        string link = xn["link"].InnerText;
        string desc = xn["description"].InnerText;
        string auth = xn["author"].InnerText;
        string pdate = xn["pubDate"].InnerText;

        XmlElement itemnode = xmlDoc.CreateElement("item");

        itemnode.InnerXml = "<title></title><link></link><description></description><author></author><pubDate></pubDate>";
        itemnode["title"].InnerText = title;
        itemnode["link"].InnerText = link;
        itemnode["description"].InnerText = desc;
        itemnode["author"].InnerText = auth;
        itemnode["pubDate"].InnerText = pdate;

        xmlDoc.DocumentElement.SelectNodes("/rss/channel")[0].AppendChild(itemnode);
    }

    // Output to screen
    xmlDoc.Save(Response.Output);

my rss feed

<?xml version="1.0" encoding="utf-8"?>
<rss xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="2.0">
<channel>
<title>My RSS Feed</title>
<link>http://www.mylink.aspx</link>
<description>
</description>
<item>
  <title>Top marks</title>
  <link>http://www.mymarks.aspx</link>
  <description>
  &lt;p&gt;description field here&lt;/p&gt;
  </description>
  <author>Viv</author>
  <pubDate>Thu, 16 Aug 2012 12:10:54 GMT</pubDate>
</item>
<item>
  <title>Costa Coffee</title>
  <link>http://www.Costa.aspx</link>
  <description>
  &lt;p&gt;Costa Coffee have special offers.&lt;/p&gt;
  </description>
  <author>Mike</author>
  <pubDate>Thu, 23 Aug 2012 15:55:53 GMT</pubDate>
</item>
<item>
  <title>Celebrate success</title>
  <link>http://www.Celebrate.aspx</link>
  <description>
  &lt;p&gt;Lets all celebrate &lt;/p&gt;
  </description>
  <author>Viv</author>
  <pubDate>Thu, 22 Aug 2012 09:10:21 GMT</pubDate>
</item>
</channel>
</rss>

Answer 1

You can do this fairly quickly and painlessly using Linq to XML.

If you parse your XML using XElement.Parse(...) you can then use OrderBy or OrderByDescending functions and alter the content pretty easily. Here is a simplified example:

XElement element = XElement.Parse(@"
<rss>
<channel>
<item title='something' pubDate='22/11/2012'/>
<item title='something2' pubDate='24/03/2012'/>
<item title='something3' pubDate='10/02/2010'/>
<item title='something4' pubDate='22/01/2011'/>
</channel>
</rss>");

var elements = element.Element("channel")
                .Elements("item")
                .OrderBy<XElement, DateTime>(e => DateTime.ParseExact(e.Attribute("pubDate").Value, "dd/MM/yyyy", null))
                .Select(e => new XElement("item",
                    new XElement("title", e.Attribute("title").Value),
                    new XElement("pubDate", e.Attribute("pubDate").Value))); // Do transform here.

            element.Element("channel").ReplaceAll(elements);

            Console.Write(element.ToString());

The XML is not going to be the same as yours, but hopefully it gives you an idea of what you could do. You can just specify XElement and XAttribute objects as content for your new XML, this outputs the following:

<rss>
  <channel>
    <item>
      <title>something3</title>
      <pubDate>10/02/2010</pubDate>
    </item>
    <item>
      <title>something4</title>
      <pubDate>22/01/2011</pubDate>
    </item>
    <item>
      <title>something2</title>
      <pubDate>24/03/2012</pubDate>
    </item>
    <item>
      <title>something</title>
      <pubDate>22/11/2012</pubDate>
    </item>
  </channel>
</rss>

I hope this is useful.