On 49 $ship_tax = rs_1['Tax']; this $ is missing
With Regards
Binoy.M.V
--- On Tue, 21/7/09, uttam takalkar <uttam.takalkar@xxxxxxxxx> wrote:
From: uttam takalkar <uttam.takalkar@xxxxxxxxx>
Subject: Re: Re: query .. please review
To: php-objects@xxxxxxxxxxxxxxx
Date: Tuesday, 21 July, 2009, 11:46 AM
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@gmail. com> wrote:
>
>
> --- In php-objects@ yahoogroups. com <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'] ;
>
>
>
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