C#: XML属性値を取得する単純な例(Linq, XPath)

gistfile1.cs

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml.Linq;
using System.Xml.XPath;
using System.Xml;

namespace Example
{
    class Program
    {
        static void Main(string[] args)
        {
            // LinqでURLを取得
            var xml = XDocument.Load("http://matome.naver.jp/feed/matome/2127198911354142701");
            var media = xml.Root.GetNamespaceOfPrefix("media");

            var query = xml.Descendants(media + "content").Select(x => x.Attribute("url").Value);
            foreach (var url in query)
            {
                Console.WriteLine(url);
            }


            // Linq + XPathでURLを取得
            var manager = new XmlNamespaceManager(new NameTable());
            manager.AddNamespace("media", media.ToString());

            var query2 = ((IEnumerable<object>)xml.XPathEvaluate("//media:content/@url", manager))
                .Select(x => ((XAttribute) x).Value);
            foreach (var url in query2)
            {
                Console.WriteLine(url);
            }
        }
    }
}