Question: is session_start() called on the page where this code is run?
var loggedin = <?=$_SESSION['loggedin']?>;
if (loggedin == true)
{
$("div#login").hide();
$("div#logout").unhide();
}
else
{
$("div#login").unhide();
$("div#logout").hide();
}
On 3/30/07, Kim Johnson <[EMAIL PROTECTED]> wrote:
I am indeed not using output buffering. I call
session_start as the first thing on every page, before
the headers are sent. I also call session_start on my
login page, first thing.
If I use output buffering, do you think that would fix
things? I could call session_start() again as a
callback function after the ajax finishes, perhaps...?
thanks,
-kim
--- Aaron Heimlich <[EMAIL PROTECTED]> wrote:
> On 3/30/07, Kim Johnson <[EMAIL PROTECTED]> wrote:
> >
> > Should I be recalling
> > session_start?
>
>
> As a general rule, you should call session_start()
> in *every* PHP script
> that accesses $_SESSION. Where you call it depends
> on whether you're using
> output buffering or not[1], but it must *always* be
> *before* you access
> $_SESSION for the first time in that script.
>
>
> [1] If you're not using output buffering (
> http://us.php.net/manual/en/ref.outcontrol.php),
> then you *must* call
> session_start() *before* any output is sent to the
> browser, otherwise you'll
> get a "can't send headers because headers were
> already sent" error (this
> means that the call to session_start() should
> probably be the very first
> thing your script does). If you are using output
> buffering, then you can
> call it pretty much anywhere in your script before
> you send the contents of
> the output buffer to the browser.
>
> --
> Aaron Heimlich
> Web Developer
> [EMAIL PROTECTED]
> http://aheimlich.freepgs.com
> > _______________________________________________
> jQuery mailing list
> discuss@jquery.com
> http://jquery.com/discuss/
>
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--
Aaron Heimlich
Web Developer
[EMAIL PROTECTED]
http://aheimlich.freepgs.com
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