xpath  查询问题
XML如下: 
  <?xml   version= "1.0 "   encoding= "utf-8 "   ?>  
  <Users>  
 	 <User>  
 		 <UserName> tyq0319 </UserName>  
 		 <UserPassword> 219875 </UserPassword>  
 		 <UserInfoUrl> UserInfo/tyq0319.xml </UserInfoUrl>  
 		 <UserMessage> l; </UserMessage>  
 		 <UserPop> ;l </UserPop>  
 	 </User>  
 	 <User>  
 		 <UserName> tyq0319 </UserName>  
 		 <UserPassword> 219875 </UserPassword>  
 		 <UserInfoUrl> UserInfo/tyq0319.xml </UserInfoUrl>  
 		 <UserMessage> l; </UserMessage>  
 		 <UserPop> ;l </UserPop>  
 	 </User>  
  </Users>      
 代码如下: 
       protected   void   Button1_Click(object   sender,   EventArgs   e) 
             { 
                         XPathNavigator   nav; 
                         XPathDocument   Xdoc   =   new   XPathDocument( "F:/WebSite2/user/UserLogin.xml "); 
                         nav=Xdoc.CreateNavigator(); 
                         this.Label1.Text   =   nav.Select( "/Users/User/UserPassword[../UserName= 'tyq0319 '] ").Current.Value; 
 }     
 为什么结果是: 
             tyq0319219875UserInfo/tyq0319.xmll;;ltyq0319219875UserInfo/tyq0319.xmll;;l   ;   
 我想根据用户名只得到密码,请问该怎么改?     
------解决方案--------------------XPathNodeIterator iterator = nav.Select( "/Users/User/UserPassword[../UserName= 'tyq0319 '] "); 
             while (iterator.MoveNext()) 
             { 
                 Console.WriteLine(iterator.Current.Value); 
             }     
 ***************************************************************************** 
 欢迎使用CSDN论坛专用阅读器 : CSDN Reader(附全部源代码)    
 最新版本:20070130   
 http://www.cnblogs.com/feiyun0112/archive/2006/09/20/509783.html
------解决方案--------------------XmlDocument doc = new XmlDocument(); 
 doc.Load( "your xml file full path "); 
 XmlNodeList users = doc.SelectNodes( "Users/User "); 
 foreach(XmlNode user in users) { 
   if(user.ChildNodes[ "UserName "].InnerText ==  "tyq0319 ") { 
     string userPassword = user.ChildNodes[ "UserPassword "].InnerText; 
   } 
 }   
 // "[] "这里面好像只能写XmlAttribute的值吧,如 <Node att1 =  "a value "/>  ,XPath =  "Node[@att1 =  'a value '] "
------解决方案--------------------像你这像的情况,可以先找出UserName,然后再找它的得到它的ParentNode,再找到它的UserPassword 
------解决方案--------------------JIEFEN 接分啊