简单的xml解析

什么是解析lat和long以下xml片段的最简单方法.没有名称空间等.

它是一个字符串变量.不是流.

      stockholm
      sweden
      
        51.1
        67.98
      

到目前为止我所阅读的所有内容都过于复杂,不应该是一项简单的任务,例如 http://geekswithblogs.net/kobush/archive/2006/04/20/75717.aspx

我一直在看上面的链接

当然在.net中有一种更简单的方法可以做到这一点？

使用Linq for XML:

   XDocument doc= XDocument.Parse("stockholmsweden51.167.98");

    var points=doc.Descendants("gpoint");

    foreach (XElement current in points)
    {
        Console.WriteLine(current.Element("lat").Value);
        Console.WriteLine(current.Element("lng").Value);
    }

    Console.ReadKey(); 

using System.IO;
using System.Xml;
using System.Xml.XPath;

...

    string xml = @"      
                     stockholm  
                     sweden
                                
                            51.1        
                            67.98    
                        
                   ";

    XmlReaderSettings set = new XmlReaderSettings();
    set.ConformanceLevel = ConformanceLevel.Fragment;

    XPathDocument doc = 
        new XPathDocument(XmlReader.Create(new StringReader(xml), set));

    XPathNavigator nav = doc.CreateNavigator();


    Console.WriteLine(nav.SelectSingleNode("/poi/gpoint/lat"));
    Console.WriteLine(nav.SelectSingleNode("/poi/gpoint/lng"));

您当然可以使用xpath SelectSingleNode将元素选择为变量.