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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