Re: Problem w/query - again

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 <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/) >> > To unsubscribe, visit: http://www.php.net/unsub.php >> > >> > >> > > Amit - > > Thanks. > > Tried it. Still does not work. > > This is the query I get: > > > select * from Intake3 where 1 > > Ethan > >