// 得到一个连接Access数据库的OleDbConnection对象DBConn
                // 以下将图片写入表T_Image的其它代码.
                oleDBConn.Open();
                System.Data.OleDb.OleDbDataAdapter dbAdapter=new OleDbDataAdapter("Select * From T_Image",oleDBConn);
                
                // 得到数据表
                DataSet dataSet=new DataSet();
                dbAdapter.Fill(dataSet,"T_Image");
                DataTable table=dataSet.Tables[0];
               
                // 读取图像文件
                System.IO.FileStream fileStream=new System.IO.FileStream(@"E:\xx\xx.jpg",System.IO.FileMode.Open,System.IO.FileAccess.Read);
                byte[] data=new byte[fileStream.Length];
                fileStream.Read(data,0,(int)fileStream.Length);
                fileStream.Close();                // 给数据行赋值
                DataRow row=table.NewRow();
                //row[0]为自动生成的值
                row[1]="图片";
                row[2]=data; //第三列的数据类型为ole对象
                table.Rows.Add(row); // 将列添加到数据表
                table.AcceptChanges();
                //到此row[2]里有数据,我用 table.DataSet.WriteXml(@"D:\xx.xml");
                // 结果文件xx.xml里面是有数据的.                //接着执行 更新,往数据库中插入数据
                int i=dbAdapter.Update(dataSet,"T_Image");
                在执行更新操作时,i值始终为0,更新操作失败,无法将数据插入数据库中,怎么回事????                // 释放
                oleDBConn.Close();
                oleDBConn.Dispose();