Problem: <?php print $pic; ?>

[Anon]

Hi,

I am using a template that dynamically calls images using php print. The code to call the image within the .php file is:

img src=<?php print $pic; ?> alt=""

(inside html tags) and then calling image into the page via the browser, I use:

template.php?$pic=1.jpg

which calls 1.jpg and so on for each image.

The problem is, the image displays in template.php as a broken image, and viewing the source shows: img src= alt=""

Is there a reason why this code doesn't display the correct image? What am I missing?

Any help greatly appreciated.

Jonathen

[Anon]

sounds like register_globals is OFF. try this:

img src=<?php print $_GET['pic']; ?>

[Anon]

I get the same thing - I checked php.ini, and register_globals is ON. Thing is, I checked this against a PHP site, and it's correct (I even removed my code and replaced with the one there in case of syntax error) which is:
http://www.phphelp.com/article/39p3.php

Am I missing a function call or something? I was wondering whether I had to tell the script what function the $pic variable uses?

shrugs

[Anon]

create a "test.php" file with the following line:

<?= $pic ?>

and then goto http://www.whatever.com/test.php?pic=this_is_a_test

what do you see?

[Anon]

I get simply:

this_is_a_test

[Anon]

it's sorted now, using the same php print $pic function - I realised how stupid I was being! I was calling the template with it's image using /template.php?$pic=1.jpg when I got up today and realised: the $ is not needed in the browser address! Instead, /template.php?pic=1.jpg works fine!

Thanks for all your help.