RE: [PHP-DB] Inserting Variable Variable?

From: Date: Wed, 18 Jul 2001 14:29:58 +0000
Subject: RE: [PHP-DB] Inserting Variable Variable?
References: 1  Groups: php.db 
Request: Send a blank email to php-db+get-10546@lists.php.net to get a copy of this message
Hi Jeff. $varname = "C_First_Name".$x; // e.g. if $x == 3 then ... echo $$varname; // ... prints the content of $C_First_Name3 Hope this is what you wanted ?! Greetinx, Mike Michael Rudel - Web-Development, Systemadministration - Besuchen Sie uns am 20. und 21. August 2001 auf der online-marketing-düsseldorf in Halle 1 Stand E 16 _______________________________________________________________ Suchtreffer AG Bleicherstraße 20 D-78467 Konstanz Germany fon: +49-(0)7531-89207-17 fax: +49-(0)7531-89207-13 e-mail: mailto:mru@suchtreffer.de internet: http://www.suchtreffer.de _______________________________________________________________ > -----Original Message----- > From: Jeff Oien [mailto:jeff@webdesigns1.com] > Sent: Wednesday, July 18, 2001 4:24 PM > To: PHP-DB > Subject: [PHP-DB] Inserting Variable Variable? > > > I want to INSERT a variable like this using MySQL. If I use this > > \"(${"C_First_Name"}.$x)\", > or > \"(${C_First_Name}.$x)\", > or > \"${C_First_Name}.$x\", > > only the $x gets interpolated in the insertion. If I use > > \"${"C_First_Name$x"}\", > > only $C_First_Name get interpolated. Any solutions? Thanks. > Jeff Oien > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-db-unsubscribe@lists.php.net > For additional commands, e-mail: php-db-help@lists.php.net > To contact the list administrators, e-mail: > php-list-admin@lists.php.net >

« previous php.db (#10546) next »