Comment: A workaround for "Notice: Only variable references should be returned by reference in"

Well, the workaround is for where it says, "return $output; - maybe it wasn't line 883 - but for some reason, PHP thinks that $output is a static reference , a constant, but when assigning the value of $output to a brand new variable, which I called $a just to keep it simple for now - it seemed to like that.

It really wasn't a problem - this error message, as it only showed up when I was in debug mode. I was just happy to find a fix for it for what is probably a weird scenerio at my host provider, freehostia.com.