Now my code is
extract ($_GET);
if ($_GET['printout'] != "yeah") { include("header.php"); }
but I still get the following error:
Undefined index: printout
I understand nothing
"Jonathan Wilkes" <[EMAIL PROTECTED]> skrev i meddelandet
news:[EMAIL PROTECTED]
Hi,
What he means is that with "register_globals=off" you cannot do this:
echo $path
you need to do this (if the variable is sent by "POST" action)
echo _POST('path')
and through "GET"
echo _GET('path')
-----Original Message-----
From: Øystein Håland [mailto:[EMAIL PROTECTED]
Sent: 03 June 2003 17:02
To: [EMAIL PROTECTED]
Subject: [PHP] Re: Migration from register_globals=on to
register_globals=off
I'm not sure what you mean. To give ONE example:
Earlier I could use this code on top of every page:
if ($printout != "yeah") { include("header.php"); }
This code gives an error today. The variable $printout is set if the visitor
choose to click on the 'print_page_image', otherwise the variable has no
value.
"Esteban FernáNdez" <[EMAIL PROTECTED]> skrev i meddelandet
news:[EMAIL PROTECTED]
> When you recivied that error ?, in a form ?, if is in a Form just put in
the
> top of .php files this code
>
> $HTTP_GET_VARS["variable2"];
> $HTTP_GET_VARS["variable3"];
>
> Of course if you send with other method (post) change the GET for POST
>
> $HTTP_POS_VARS["variable2"];
> $HTTP_POS_VARS["variable3"];
>
> Regards.
>
> Esteban.
>
>
>
> "ØYstein HåLand" <[EMAIL PROTECTED]> escribió en el mensaje
> news:[EMAIL PROTECTED]
> > None of my old scripts worx nowadays and the most common error message
is
> > 'undefined variable'. What is the best/simplest way to work around this
> > situation?
> > if !isset($myvar) {
> > do this
> > blah blah
> > }
> > ?
> >
> >
>
>
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