Try this:
<form name="myEditForm" action="edit.phtml?number=<?php echo $exedit; ?>"
method="post" onsubmit="return validate()">
This should display ?number correctly (I take it that's the problem).
You may also need to use "return validate();" or "javascript: return
validate();", not too sure about this though.
HTH
Jon
-----Original Message-----
From: Curtis [mailto:[EMAIL PROTECTED]]
Sent: 09 April 2001 17:09
To: php
Subject: [PHP] Passing variables,,, I know its not this hard.
Hello,
Could someone please tell me where I am going wrong here.
I have an HTML form and I want to past the input from a text box to the
next page url to connect
to my mysql database.
Here is where I am at now....
<form name="myEditForm" action="edit.phtml?number=<? '$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
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