Because it's not $s in the context of a printf() call. It is %s to denote a
string replacement.
-Mike
- Original Message -
From: "Matthew Crouch" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, June 24, 2002 12:41 PM
Subject: [PHP-DB] got it,
Seth Yount wrote:
> in you print statement you have ?lastname=%s, shouldn't that be '$s'
> denoting a variable being passed? Try that out.
>
> gl -- Seth
>
> Matthew Crouch wrote:
>
> > this bit from my index page is giving me 2 headaches:
> > 1. it isn't passing anything into the URL
> > 2. t
