RE: [PHP] Passing variables,,, I know its not this hard.

Try this: <form name="myEditForm" action="edit.phtml?number=<?php echo $exedit; ?>" method="post" onsubmit="return validate()"> This should display ?number correctly (I take it that's the problem). You may also need to use "return validate();" or "javascript: return validate();", not too sure about this though. HTH Jon -----Original Message----- From: Curtis [mailto:curtis_c@bellsouth.net] Sent: 09 April 2001 17:09 To: php Subject: [PHP] Passing variables,,, I know its not this hard. Hello, Could someone please tell me where I am going wrong here. I have an HTML form and I want to past the input from a text box to the next page url to connect to my mysql database. Here is where I am at now.... <form name="myEditForm" action="edit.phtml?number=<? '$exedit' ?>" method="post" onsubmit="return validate()"> <input type="text" name="exedit" size="20"> <input type="submit" value="Submit" name="submit"> </form> the onsubmit is a javascript for validation. I have tried this every which way, but the correct way. Could someone point me in the right direction. Thanks Curtis -- 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 ********************************************************************** 'The information included in this Email is of a confidential nature and is intended only for the addressee. If you are not the intended addressee, any disclosure, copying or distribution by you is prohibited and may be unlawful. Disclosure to any party other than the addressee, whether inadvertent or otherwise is not intended to waive privilege or confidentiality' **********************************************************************