try
            {                string strappdir = Path.GetDirectoryName(System.Reflection.Assembly.GetExecutingAssembly().GetName().CodeBase);
                string infofile = Path.Combine(strappdir, "data.xml");
                XmlDocument doc = new XmlDocument();
                doc.Load(infofile);
                 Cursor.Current = Cursors.WaitCursor;
                 XmlNodeReader reader = new XmlNodeReader(doc);
                 DataSet ds = new DataSet();
                 ds.ReadXml(reader);
                 reader.Close();
                 DataTable dt = ds.Tables[0];
                 this.dataGrid1.DataSource = dt.DefaultView;
                Cursor.Current = Cursors.Default;
            }
            catch(XmlException e)
            {
                MessageBox.Show(e.Message);
            }
报找不文件,请问代码要如何写还能把xml获取出来呢,请高手帮帮忙啊