How to pick a record in XML file

Hi..

Try This Code It May Help You
 protected void btnSearch_Click(object sender, EventArgs e)
    {
             string empid1 = txtEmpid.Text.Trim() ;
            XmlDocument doc = new XmlDocument();
            string xmlFile = System.Web.HttpContext.Current.Server.MapPath("~/App_Data/employee.xml");
            doc.Load(xmlFile);
            int empid = int.Parse(empid1);
            XmlNodeList xmlnode = doc.GetElementsByTagName("employee");
            for (int i = 0; i < xmlnode.Count; i++)
            {
                XmlAttributeCollection xmlattrc = xmlnode[i].Attributes;
                if (int.Parse(xmlattrc[0].Value)  > empid)
                {
                    txtEmpid.Text = xmlattrc[0].Value;
                    txtFname.Text = xmlnode[i].ChildNodes[0].InnerText; //Assign First Name from XML file to txtFirstName Text Box
                    txtLname.Text = xmlnode[i].ChildNodes[1].InnerText; //Assign Last Name from XML file to txtLastName Text Box
                    txtAge.Text = xmlnode[i].ChildNodes[2].InnerText;// //Assign Age from XML file to txtAge Text Box
                    return;
                }
                else
                {
                    //txtEmpid.Text = xmlattrc[0].Value; ;
                    txtFname.Text = ""; //Assign First Name from XML file to txtFirstName Text Box
                    txtLname.Text = ""; //Assign Last Name from XML file to txtLastName Text Box
                    txtAge.Text = "";// //Assign Age from XML file to txtAge Text Box
                    lblMsg.Text = "No record Found. !";
                }
            }
    }
Thank's
Prasanta Kumar Pradhan
http://globaltouch.in