c#读取xml文件到datagridview实例

namespace QueryXMLByLINQ
{
    public partial class Frm_Main : Form
    {
        public Frm_Main()
        {
            InitializeComponent();
        }

        static string strPath = "Employee.xml";
        static string strID = "";

        //窗体加载时加载XML文件
        private void Form1_Load(object sender, EventArgs e)
        {
            getXmlInfo();
        }

        //显示选中XML节点的详细信息
        private void dataGridView1_CellClick(object sender, DataGridViewCellEventArgs e)
        {
            strID = dataGridView1.Rows[e.RowIndex].Cells[3].Value.ToString();//记录选择的职工编号
            XElement xe = XElement.Load(strPath);//加载XML文件
            //使用LINT从XML文件中查询信息
            IEnumerable<XElement> elements = from PInfo in xe.Elements("People")
                                             where PInfo.Attribute("ID").Value == strID
                                             select PInfo;
            foreach (XElement element in elements)//遍历查询结果
            {
                textBox11.Text = element.Element("Name").Value;//显示职工姓名
                comboBox1.SelectedItem = element.Element("Sex").Value;//显示职工性别
                textBox12.Text = element.Element("Salary").Value;//显示职工薪水
            }
        }

        #region 将XML文件内容绑定到DataGridView控件
        /// <summary>
        /// 将XML文件内容绑定到DataGridView控件
        /// </summary>
        private void getXmlInfo()
        {
            DataSet myds = new DataSet();
            myds.ReadXml(strPath);
            dataGridView1.DataSource = myds.Tables[0];
        }
        #endregion
    }
}