Re: Problem w/query - again

At 12:48 AM 2/10/2012, Amit Tandon wrote:

Dear Ethan The line you are getting is because the $_POST[fieldname] is blank. So for the following line  if ( ! empty( $_POST['field'] ) ) change it to  if ( ! empty( $_POST["$field"] ) ) Your line : Program is searxchinbg for variable name field New line : The Program is seacging for varable stored in $field. Rember to use double quotes And to verify the value echo $_POST["$field"] before your if line i.e.  if ( ! empty( $_POST["$field"] ) )    ============ regds amit "The difference between fiction and reality? Fiction has to make sense." On Fri, Feb 10, 2012 at 11:04 AM, Ethan Rosenberg <<mailto:ethros@earthlink.net>ethros@earthlink.net> wrote: At 12:13 AM 2/10/2012, Amit Tandon wrote: Dear Ethan It seems you are trying to build a query.But you are not getting field names. If you required field names then change the following line to foreach ( $allowed_fields AS $field => $_POST['field']) to foreach ( $allowed_fields AS $field) This would convert the variable field to value. In yoyr line the variable field is treated as array index ============ regds amit "The difference between fiction and reality? Fiction has to make sense." <snip> Advice and help please. Thanks. Ethan PHP Database Mailing List (<http://www.php.net/>http://www.php.net/) To unsubscribe, visit: <http://www.php.net/unsub.php>http://www.php.net/unsub.php

Amit - Thanks. Tried your edit. Still does not work. Ethan

Amit - Thanks. Tried it.  Still does not work. This is the query I get: select * from Intake3 where  1 Ethan