note 97197 rejected from function.imagegif by danbrown

Note Submitter: Kris dot Craig at gmail dot com ---- It should be noted that, contrary to what the manual implies, an error will be returned if you specify a NULL filename instead of simply omitting it. For example: function Test( $img, $filename = NULL ) { $ok = imagegif( $img, $filename ); } This will return an image that contains errors. However, if you simply omit $filename altogether, you'll be just fine. To make the above function work error-free, do this: function Test( $img, $filename = NULL ) { if ( $filename == NULL ) { $ok = imagegif( $img ); } else { $ok = imagegif( $img, $filename ); } } The manual somewhat inaccurately states the following for the $filename argument: "If not set or NULL, the raw image stream will be outputted directly." In fact, as I just demonstrated, this is misleading as it fails to mention that passing it as a NULL value causes a broken (error) image to be displayed. I would recommend you avoid this by simply making sure that you don't pass a NULL value for $filename.