Question

I have a link that recieves a dyanamic image via php, and when you click it, it opens the image in a new window. It looks like this:

<a href="{IMAGE_FULL_URL}" target="_blank">Click Here</a>

I want to open the image on a page called red.php, and inside a div with the id=green.
From the sending page I’m thinking of something like this:

<a href="red.php?{IMAGE_FULL_URL}">Click Here</a>

On red.php the code would possibly be something like this:

<?php

   $picture = $_GET['IMAGE_FULL_URL'];
   echo ".$picture"
?>

I’m sure I have something wrong above, and I don’t know where to begin to add the div id to the url, or if I can simply place the php code inside the div. Can someone please show me the code so this will work? Thanks.

Answer 1

here is an example:

your initial HTML:

<a href="path/to/red.php?pic=<?php echo urlencode('path/to/image'); ?>">Click Here</a>

your Red.php:

<div id="green">
  <img src="<?php echo urldecode($_GET['pic']); ?>" alt="" />
</div>