如何获取XML文件节点相关列表？....

Question

本帖最后由 O7110650504 于 2011-08-08 10:04:07 编辑

Answer 1

<?xml version="1.0" encoding="utf-8"?>
<Resp>
  <H>
    <Ver>版本号</Ver>
    <Sc>状态码</Sc>
    <Msg>返回消息</Msg>
  </H>
</Resp>
其次，你可以用一下代码取得你想要的值：
public static void HandleXML()
        {
            XmlDocument xmldoc = new XmlDocument();
            xmldoc.Load("C:\\11.xml");
            XmlNode xnRoot = xmldoc.SelectSingleNode("Resp");
            XmlNodeList nodelist = xnRoot.ChildNodes;
            foreach (XmlNode node in nodelist)
            {
                string Msg=node.SelectSingleNode("Msg").InnerXml;
                if (Msg !="")
                {
                    //在此Msg 就是你想要的值
                     Console.WriteLine(Msg);
                }
            }
        }

Answer 2

            string path = @"E:\x.xml";
            XmlDocument xml = new XmlDocument();
            xml.Load(path);
            XmlNodeList node= xml.SelectNodes("//name");
            foreach (XmlNode n in node)
                Console.WriteLine(n.InnerText);
            Console.ReadLine();

Answer 3

    public XmlDocument xml = new XmlDocument(); string path = @"E:";
            string[] file = Directory.GetFileSystemEntries(path);
            dgvPathWay.Rows.Clear();
            for (int i = 0; i < file.Length; i++)
            {
                xml.Load(file[i]);
                XmlElement root = xml.DocumentElement;
                DataGridView view = new DataGridView();
                int index = dgvPathWay.Rows.Add(view);
                dgvPathWay.Rows[index].Cells["name"].Value = root.Name;
                dgvPathWay.Rows[index].Cells["dept"].Value = root.GetAttribute("");
            }