求教高手 ： C# XML SelectSingleNode的问题

Question

XmlDocument doc = new XmlDocument();
             doc.Load("LanguagePackage.xml");
             XmlElement root = null, theGroup = null, theItems = null, theItem = null;
             root = doc.DocumentElement;
             theGroup = (XmlElement)root.SelectSingleNode("/Groups/Group[GroupName='PanSchemaClientControl']");
得到的theGroup 总是为空，（root不为空）求帮助！下面是我的xml文件的内容
<?xml version="1.0"?>
<LanguagePackage Version="1.0">
     <Groups>
         <Group GroupName="PanSchemaClientControl" Description="平台客户端PanSchemaClientControl.dll的语言包">
         </Group>        
     </Groups>
</LanguagePackage>

Answer 1

        static void Main(string[] args)
        {
            XmlDocument doc = new XmlDocument();
            doc.Load("XMLFile1.xml");
            XmlElement root = null, theGroup = null, theItems = null, theItem = null;
            root = doc.DocumentElement;
            
            theGroup = (XmlElement)root.SelectSingleNode("Groups");
            theItems = (XmlElement)theGroup.SelectSingleNode("Group");            Console.WriteLine("根节点是:"+theGroup.Name);
            Console.WriteLine("次节点是:" + theItems.Name);
            Console.WriteLine("次节点属性:");
            foreach (XmlAttribute a in theItems.Attributes)
            {
                Console.WriteLine(a.InnerXml);
            }
            Console.ReadLine();
        }

Answer 2

XmlDocument doc= new XmlDocument();
doc.Load("");
XmlNode no= doc.SelectSingleNode("LanguagePackage/Groups/Group[GroupName='PanSchemaClientControl']");   
if(no!=null)
{}

Answer 3

XmlNode node = root.SelectSingleNode("//Group[. ='PanSchemaClientControl']");

Answer 4

LanguagePackage/Groups/Group[GroupName='PanSchemaClientControl']

Answer 5

theGroup = (XmlElement)root.SelectSingleNode("/Groups/Group[@GroupName='PanSchemaClientControl']");没有记错的话，xpath里面，[]里面的参数前面是必须加@的你去验证一下，也只有个大概的印象了

Answer 6

mlDocument doc = new XmlDocument();
  doc.Load("LanguagePackage.xml");
应该用doc.LoadXml("LanguagePackage.xml") 这个方法例子：
    XmlDocument xmlDoc = new XmlDocument();
            xmlDoc.LoadXml(strXmlPath);
            XmlNode     root          = xmlDoc.DocumentElement;
            XmlNodeList xlName        = xmlDoc.GetElementsByTagName( "Name");
            if ( xlName.Count>0 )
                this.m_strName = xlName[0].InnerText;