workParm = mySqlCommand.Parameters.Add( "@a_TableList ", SqlDbType.VarChar, 200);
mySqlCommand.Parameters[ "@a_TableList "].Value = "OFFERID,type,offertime ";
这是处理sql数据库的，要处理access数据库的应该怎么做，请大家帮忙
------解决方案--------------------------------------------------------
System.Data.OleDb.OleDbParameter paramPersonName = new OleDbParameter( "@PersonName ", System.Data.OleDb.OleDbType.VarChar,50);
paramPersonName.Value = fileTitle;
command.Parameters.Add(paramPersonName);