Undefined index:

I'm using this code to check if a cookie has a value :

If(!IsSet($HTTP_COOKIE_VARS["LoggedIn"])) ....


In case that this cookie is not set, how can i avoid the
Undefined index: line in the error log?

berber

At 2:37 PM +0200 6/16/07, WeberSites LTD wrote:
>I'm using this code to check if a cookie has a value :
>
> If(!IsSet($HTTP_COOKIE_VARS["LoggedIn"])) ....
>
>
>In case that this cookie is not set, how can i avoid the
>Undefined index: line in the error log?
>
>berber

berber:

Try this:

$loggedIn = isset($HTTP_COOKIE_VARS["LoggedIn"]) ?
($HTTP_COOKIE_VARS["LoggedIn"]) : "not logged in";
echo($loggedIn);

Cheers,

tedd
--
-------
http://sperling.com http://ancientstones.com http://earthstones.com

WeberSites LTD wrote:
> I'm using this code to check if a cookie has a value :
>
> If(!IsSet($HTTP_COOKIE_VARS["LoggedIn"])) ....
>
>
> In case that this cookie is not set, how can i avoid the
> Undefined index: line in the error log?

The isset function does not give an error if you pass it a non-existant
variable. Are you sure it's this line that's causing the undefined index
error?

-Stut

There is obviously something else wrong. That's the purpose of isset();

Comment out your line and see what happens.

I'd guess the real problem is later in your code when you are trying to do
something with $HTTP_COOKIE_VARS["LoggedIn"] even though your isset() test said
it didn't exist.

WeberSites LTD wrote:
> I'm using this code to check if a cookie has a value :
>
> If(!IsSet($HTTP_COOKIE_VARS["LoggedIn"])) ....
>
>
> In case that this cookie is not set, how can i avoid the
> Undefined index: line in the error log?
>
> berber
>