I can't see anything sticking out...but why have you done this...
$test2Url = "test2.php?PHPSESSID=" . session_id( );
You don't need to do this. Try doing this
$test2Url = "test2.php";
Your server might be set up to use a different name for sessions
John Wards
SportNetwork.net
----- Original Message -----
From: "Chris" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, September 11, 2002 9:47 AM
Subject: [PHP] sessions nightmare
> Greetings,
>
> I am a new user to php and sessions and for the life of me I
> cannot figure this out....
>
> I want to be able to have a user login (which is completed), they goto a
> search page (completed), and search for a particular item (completed). A
> page will display all the links for the items searched for (completed).
> What I want out of sessions is to when they click on the link for a
> particular item, the item number stay in a session so it can be called
> through out each page they goto. What I have as a base of test code is the
> following (this was taken from someone's example):
>
> test1.php:
>
> <?php
> session_start( );
> session_register("SESSION");
>
> *** THE FOLLOWING VARIABLE ($retrived_itemno) WOULD NEED TO BE SET ON THE
> *** SEARCH PAGE ***
> $SESSION["item"] = $retrived_itemno;
>
> $test2Url = "test2.php?PHPSESSID=" . session_id( );
>
> ?>
>
> <a href="<?=$test2Url ?>">Goto next page</a>
>
>
> test2.php:
>
> <?php
> session_start( );
> echo "the retrived value equals: $SESSION[item]";
> ?>
>
>
>
> After you click the hyperlink on test1.php, test2.php will load and the
> session ID is in the URL, but nothing is displayed in the echo for
> $SESSION[item] in test2.php. What is going on?!#!#!$#
>
>
> Thanks a million in advance!!!!!!!!!!!!!!!!!!
> - Chris
>
>
>
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