问题描述

asp.net 如何读取 这样的XML 取得它的loginname和realname

 <cas:serviceResponse xmlns:cas='http://www.yale.edu/tp/cas'>
<cas:authenticationSuccess>
        <cas:user>bb</cas:user>
        <cas:attributes  nid='1'>
        <cas:id>2</cas:id>
        <cas:loginname>bb</cas:loginname>
        <cas:role>2</cas:role>
        <cas:realname>bbggg</cas:realname>
        </cas:attributes>
    </cas:authenticationSuccess>
</cas:serviceResponse>

解决方案

```void Main()
{
string xml = @"

cas:authenticationSuccess

cas:userbb/cas:user

cas:id2/cas:id

cas:loginnamebb/cas:loginname

cas:role2/cas:role

cas:realnamebbggg/cas:realname

/cas:attributes

/cas:authenticationSuccess

/cas:serviceResponse";

XDocument xmlPage = XDocument.Parse(xml);

XNamespace ns = "http://www.xxx.edu/tp/cas";

       var query = from x in xmlPage.Descendants(ns+"authenticationSuccess")
                   from y in x.Element(ns+"attributes").Elements()
                   select y;
       foreach(XElement item in query)
            Console.WriteLine("{0}:t  {1}",item.Name.LocalName,item.Value);

     /*
     id:      2
    loginname:    bb
    role:     2
    realname:     bbggg
     */

}

解决方案二：

 void Main()
{
    string xml = @" <cas:serviceResponse xmlns:cas='http://www.yale.edu/tp/cas'>
<cas:authenticationSuccess>
        <cas:user>bb</cas:user>
        <cas:attributes  nid='1'>
        <cas:id>2</cas:id>
        <cas:loginname>bb</cas:loginname>
        <cas:role>2</cas:role>
        <cas:realname>bbggg</cas:realname>
        </cas:attributes>
    </cas:authenticationSuccess>
</cas:serviceResponse>";
           XDocument xmlPage = XDocument.Parse(xml);
           XNamespace ns = "http://www.yale.edu/tp/cas";  

           var query = from x in xmlPage.Descendants(ns+"authenticationSuccess")
                       select new
                        {
                            loginname = x.Element(ns+"attributes").Element(ns+"loginname").Value,
                            realname = x.Element(ns+"attributes").Element(ns+"realname").Value
                        };  

}

解决方案三：

你的问题主要还是在 XNamespace

解决方案四：

如果attatributes下节目未知，直接这样：

  var query = from x in xmlPage.Descendants(ns+"authenticationSuccess")
                       select x.Element(ns+"attributes");

解决方案五：

参考这个：

 void Main()
{
    string xml = @" <cas:serviceResponse xmlns:cas='http://www.xxx.edu/tp/cas'>
<cas:authenticationSuccess>
        <cas:user>bb</cas:user>
        <cas:attributes  nid='1'>
            <cas:id>2</cas:id>
            <cas:loginname>bb</cas:loginname>
            <cas:role>2</cas:role>
            <cas:realname>bbggg</cas:realname>
        </cas:attributes>
    </cas:authenticationSuccess>
</cas:serviceResponse>";
           XDocument xmlPage = XDocument.Parse(xml);
           XNamespace ns = "http://www.xxx.edu/tp/cas";    

           var query = from x in xmlPage.Descendants(ns+"authenticationSuccess")
                       from y in x.Element(ns+"attributes").Elements()
                       select y;
           foreach(XElement item in query)
                Console.WriteLine("{0}:t  {1}",item.Name.LocalName,item.Value);

         /*
         id:      2
        loginname:    bb
        role:     2
        realname:     bbggg
         */
}