Re: Evaluting form variables - I don't see what's wrong with code

if you want to use printf, you gotta use %. eg, printf ("blah %s", $var) what follwos after the % is the same as in C or C++. ---------------- remco chang bountyquest.com On Mon, 14 Aug 2000, J. Myers wrote: > Hello Folks! This is probably the simplest problem but we're stumped. > We've set up a simple HTML form and now would like to evaluate the > submitted form variables. > > Eg. > > Form variable name: <input type="text" name="D0_0600SHOW"> > > Script: printf("The value is", $D0_0600SHOW); > > Output: "The value is " > > It's as if the variable doesn't exist. I've sent the same form variable > to another scripting language that I use and it works just fine. > What could I be doing wrong? > > Thanks ahead of time! > > Jr. > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net > For additional commands, e-mail: php-general-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net > >