Because it's not $s in the context of a printf() call. It is %s to denote a
string replacement.
-Mike
----- Original Message -----
From: "Matthew Crouch" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, June 24, 2002 12:41 PM
Subject: [PHP-DB] got it, thanks
>
>
> Seth Yount wrote:
>
> > in you print statement you have ?lastname=%s, shouldn't that be '$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. the page that gets called ("name.php") sits and tries to load
> > > forever. it looks like it's even filling up my hard drive with data (?!)
> > >
> > > Note: it does this (#2) even if I type in the URL with a variable and
> > > value, like "name.php?lastname=smith"
> > >
> > > <?
> > > printf ("<form method=\"post\" action=\"name.php?lastname=%s\">",
> > > $lastname);
> > > ?>
> > > <div align="center">
> > > <input type="Text" name="lastname" size="75">
> > > <input type="Submit" name="submit" value="Search for Last Name">
> > > </div>
> > > </form>
> > > </div>
> > >
>
>
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