How to parse XML with namespace

Question

Bonjour,

I have an xml doc:

<ns2:feeds xmlns:ns2="XXXX" xmlns="XXXXX" version="3.0">
    <ns2:feed>
        <name>XXX</name>
    </ns2:feed>
    <ns2:feed>
        <name>XXX</name>
    </ns2:feed>
    <ns2:feed>
        <name>XXX</name>
    </ns2:feed>
</ns2:feeds>

How can I use LinqToXml to get a list of Name properties? nothing I try seems to work...

var doc = XDocument.Load(@"feed.xml");

var names = doc
.XPathSelectElements("/*/*[localname()='feeds']") //What should the Xpath be, here?
.Select(p => new 
{ 
    Name = p.Descendants("name").First().Value
})
.ToList();

Is there a simple way to achieve this?

Does this work? XNamespace ns = "XXXXX"; and then p.Descendants(ns + "name").First().Value. — Corak
– Corak, Commented Aug 14, 2013 at 11:42

keyboardP · Accepted Answer · 2013-08-14 12:37:48Z

8

You can do this

XNamespace ns = XNamespace.Get("XXXX");
var listOfNames = doc.Descendants(ns + "feed")
                     .Select(x => x.Elements().First().Value).ToList();

+1 for lazyberezovsky's answer. If you need to specify the element name (name in this case) or you could have multiple name elements then you need to add a second namespace for those elements.

XNamespace ns2 = XNamespace.Get("XXXXX");
var listOfNames = doc.Root.Descendants(ns2 + "name").Select(x => x.Value).ToList();

edited Aug 14, 2013 at 12:37

answered Aug 14, 2013 at 11:45

keyboardP

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4 Comments

Henk Holterman Over a year ago

XNamespace.Get() doesn't seem to be the best approach here.

keyboardP Over a year ago

Usually with namespaces I've used the above but maybe it hasn't always been efficient in that case. What would be the alternate way?

Christopher Stevenson Over a year ago

If you are really concerned about efficiency, you could parse the xml with XmlReader.

keyboardP Over a year ago

@ChristopherStevenson - Not too concerned about efficiency in performance terms but in the way that XNameSpace.Get not being the best approach.

Sergey Berezovskiy · Accepted Answer · 2013-08-14 12:11:32Z

6

With XPathSelectElements you should provide namespace manager in order to use namespaces in XPath query:

var manager = new XmlNamespaceManager(new NameTable());
manager.AddNamespace("ns2", "XXXX");
manager.AddNamespace("ns", "XXXXX"); // default namespace

var names = from n in xdoc.XPathSelectElements("//ns2:feed/ns:name", manager)
            select (string)n;

Without XPath you should use XNamespace when providing name of node to find:

XNamespace ns = "XXXXX";
XNamespace ns2 = "XXXX";

var names = from f in xdoc.Descendants(ns2 + "feed")
            select (string)f.Element(ns + "name");

answered Aug 14, 2013 at 12:11

Sergey Berezovskiy

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Comments

Cylian · Accepted Answer · 2013-08-14 13:10:07Z

To get rid of namespaces in XLinQ queries, use similar method mentioned below:

String xml_string = @"<ns2:feeds xmlns:ns2=""XXXX"" xmlns=""XXXXX"" version=""3.0"">
                          <ns2:feed>
                              <name>XXX</name>
                          </ns2:feed>
                          <ns2:feed>
                              <name>YYY</name>
                          </ns2:feed>
                          <ns2:feed>
                              <name>ZZZ</name>
                          </ns2:feed>
                      </ns2:feeds>";

var query = XElement.Parse(xml_string).Descendants()
           .Where(c => c.Name.LocalName.ToString() == "name")
           .ToArray();

foreach (String item in query)
{
    Console.WriteLine(item);
}

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