Hello just u r not concat the variablle test this one $qry = "SELECT * FROM cityrates WHERE id=.'$city_rec'"; your code-------- $qry = "SELECT * FROM cityrates WHERE id = '$city_rec'"; --- In php-objects@xxxxxxxxxxxxxxx, "musika_astig" <mjdiose@...> wrote: > > --- In php-objects@xxxxxxxxxxxxxxx, Reji Jayan <for_rejijayan@> wrote: > > > > 44 $qry = "SELECT * FROM cityrates WHERE id = '$city_rec'"; > > 45 $rec1 = mysql_query($qry); > > 46 $rs1=mysql_fetch_array($rec1); > > 47 > > 48 $ship_county=$rs1['County']; > > 49 $ship_tax = rs_1['Tax']; > > 50 $ship_city=$rs_1['City']; > > > > got .. and im getting the error ...... > > > > Parse error: parse error in > > E:\wamp\www\blah-bhal\vieworders.php on line 49 > > > > > > i dont know wht hapnd to that line ...... > > > > is it any reserved word .. any solution to solve this > > > > ( not on renaming the field on the table .. if so) > > > > > > -------------------------------------------------------------------- > > > > > > > > > > Looking for local information? Find it on Yahoo! Local http://in.local.yahoo.com/ > > > > [Non-text portions of this message have been removed] > > > > > > it think you forgot to write a dollar sign > something like this -> $ship_tax = $rs_1['Tax']; >
