I want to include a php page if the url as the variable ?logout
My objective is to show a page (logout-message.php) once user logouts (he goes automatically to mysite.com/?logout)
Please note that the logout-message.php page contains a frame, so want I want is to display that frame over the index page (mysite.com/).
Here's what I have:
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- <?php
- if(isset($_GET['logout']))
- include("logout-message.php");
- ?>
The problem I'm encountering is that the php include seems to take away all the rest of the page. When i navigate to mysite.com/?logout I see only the content of the logout-message.php page, the content of the index page is gone!
I'd appreciate your help, thanks in advance.
If you know of an easier way to do want I want, like i.e. something like php if .... (same as above) .. display content of another page.. i don't know... i'm a newbie on php..
What I want is to display the content of an external page (over the already existing index page) if the url has ?logout.
thank You.