Podcasted filesA workaround for "Notice: Only variable references should be returned by reference in"
Posted by Simplify3 on Jul. 2 2007, commented by Simplify3 on Jul. 4 2007, (popular)
Rate this page
Get this error when using verbose output? Try this.
An example is in UTF8.php.At the end, where it says: return $output; [line 883]
change it to this:
//return $output; [comment it out]
and add:
$a=$output; return a$;
This turns it into a variable reference and it stops complaining.
Thoughts?
Bernard from nearby-an-airport Associate 6406 posts registered on Sep. 12 2003 |
The general approach is ok, but in my opinion this does not apply to the line 883 of shared/utf8.php, since $output is set a couple of lines above... Maybe at other locations?
|
Simplify3 Member 44 posts registered on June 3 2007 |
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. |
Reply
Quote
