Xml读取方法

Question

<?xml version="1.0" encoding="utf-8" ?>
<Student>
  <Name>周杰杰</Name>
  <Age>20</Age>
  <Hobby>耍耍双节棍棍</Hobby>
  <Name>张靓靓</Name>
  <Age>25</Age>
  <Hobby>唱歌歌</Hobby>
</Student>
怎么以<Name>为父节点<Age>、<Hobby>为子节点读取到Treeview中?

Answer 1

http://apps.hi.baidu.com/share/detail/298616

Answer 2

System.Xml.Linq.XDocument doc = new XDocument(yourFileName);
TreeNode nodeRoot = new TreeNode(doc.Element("Student").Element("Name").InnerText;
TreeNode nodeAge = new TreeNode(doc.Element("Student").Element("Age").InnerText;
TreeNode nodeHobby = new TreeNode(doc.Element("Student").Element("Hobby").InnerText;
nodeRoot.Nodes.Add(nodeAge);
nodeRoot.Nodes.Add(nodeHobby);
myTreeView.Nodes.Add(nodeRoot);

Answer 3

        XmlDocument xml = new XmlDocument();
        xml.Load(Server.MapPath("xml/XMLFIle.xml"));
        XmlNodeList node = xml.SelectNodes("Student/Name");
        foreach (XmlNode n in node)
        {
            TreeNode parent = new TreeNode(n.InnerText, n.InnerText);
            XmlNode first =  n.NextSibling;
            XmlNode second = first.NextSibling;
            parent.ChildNodes.Add(new TreeNode(first.InnerText, first.InnerText));
            parent.ChildNodes.Add(new TreeNode(second.InnerText, second.InnerText));
            TreeView1.Nodes.Add(parent);
        }

Answer 4

如果楼主能改这个结构,建议楼主还是改一下
例如:<?xml version="1.0" encoding="utf-8" ?>
<Students>
 <Student>
  <Name>周杰杰</Name>
  <Age>20</Age>
  <Hobby>耍耍双节棍棍</Hobby>
 </Student>
 <Student>
  <Name>张靓靓</Name>
  <Age>25</Age>
  <Hobby>唱歌歌</Hobby>
 </Student>
</Students>

Answer 5

 private void InitXMLpath()
    {
        string xmlpath = "~/aaa.Xml";
        
        _XMLpath = HostingEnvironment.MapPath(xmlpath);
        FileIOPermission permission = new FileIOPermission(FileIOPermissionAccess.Write, _XMLpath);
        permission.Demand();
    }    private DataTable GetAllNodes()
    {
        DataSet ds = new DataSet();
        ds.ReadXml(_XMLpath);
        return ds.Tables[0];
    }
//然后读取数据集
//数据集应该会读吧