RE: [PHP] Evaluting form variables

Is there a specific reason why you are using printf. You can simply use print "The value is $D0_0600SHOW"; -----Original Message----- From: J. Myers [mailto:jr@mediastop.com] Sent: Monday, August 14, 2000 8:29 AM To: php-general@lists.php.net Subject: [PHP] Evaluting form variables 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.