[Resolved] Getting a url? Almost like $PHP_SELF?

Black_Hobbit

Hi,

I have a php script that outputs a png file via the gd. Everything works fine except that someday I want to be able to display the site that a user is looking at from my file. (assuming that the user is not just looking a the file directly) I've tried just using the variable $PHP_SELF but that doesn't work.

[more about what I mean]

Well, suppose that the [img] tag was inabled on this forum. Then, I could have my image show up on this page. Well because it was on this page, the text on the image would show:

http://www.phpbuilder.com/board/newthread.php?s=&action=newthread&forumid=10

next I need to be able to separate all this out to get just:

www.phpbuilder.com

How can this be accomplished?

drew010

if the user's browser passes it, the $_SERVER['HTTP_REFERER'] variable should show the page they are viewing since the img call comes from that page.
here is an example i just made on my server. testing.php contains the image. test.php is making the call.

html page w/ img src: http://www.drew-phillips.com/gd/test.php
gd image: http://www.drew-phillips.com/gd/testing.php

then you can capture the server using string functions or regular expressions.

LordShryku

[man]parse_url[/man]

[root@geeklab]# php -r '
$t="http://www.phpbuilder.com/board/newthread.php?s=&action=newthread&forumid=10";
print_r(parse_url($t));'
Array
(
    [scheme] => http
    [host] => www.phpbuilder.com
    [path] => /board/newthread.php
    [query] => s=&action=newthread&forumid=10
)

Black_Hobbit

Wow! thanks, both of you. I can't beleive you went though all that trouble to make that example!

It works perfectly for me.