Get page url for original document when using an include

[Dogstar]

I am trying to add a 'feedback' button on every page on a website. I want to add the button to the page footer which is accessed on each page using the include function. I want the feedback button to open a email form with a hidden option to include the url that the user has come from. I have tried

$url="http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];

and

$url=$HTTP_SERVER_VARS['SCRIPT_FILENAME'];

both of which give me the footer url rather than the actual page which the footer is included in.

Any Ideas?




In case the above doesn't make sense I have outlined the basics of the page coding below

html
head
title etc
/head
body
contents of page
php include footer.php
/body
/html

and the php code I have attempted is in the footer.php page

[harshmaul]

$url="http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];
Gives the current page.....


$referer = $_SERVER['HTTP_REFERER'];
Gives where the user clicked through from.....

I'm not sure what you mean but i use the above code to track users through my sites.

[Dogstar]

Thanks...FInally got it working. Sorry for bad explanation of the problem.

T