You fetched result in $rs1 variable and you are using $rs_1. so check it properlly On Tue, Jul 21, 2009 at 11:27 AM, musika_astig <mjdiose@xxxxxxxxx> wrote: > > > --- In php-objects@xxxxxxxxxxxxxxx <php-objects%40yahoogroups.com>, 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']; > > > [Non-text portions of this message have been removed]
