I had assumed this should work, but it doesn't. am I doing something wrong or do is the only way to do this to use the eval() statement?
<?php
$type = 'today';
$todayimage = '/images/today.gif';
echo '<img src="'. ${$type}image .'" />';
?>
obviously, $type and $todayimage are defined elsewhere... and there are more $type possibilites and a $typeimage for each $type...
so do I have to do eval('echo $type,"image"'); ?
lol... that eval won't work... I'll figure out the right one now...