On Sat, 15 Jan 2005, Steven Simmons wrote: > I'm trying to write an object oriented system for saving my uploaded > files to a blog. > > I am using php to resize the original photos and save a thumbnail . > > When I get all done, I want to do something like this: > > $tempImageData = imagejpeg( $imageResource, '', 75 ); > $outputHandler -> save ( $tempImageData ); > > Where the $outputHandler is an object that knows how/where to save the file. > However, I've noticed that imagejpeg() only returns a number 1. > > This worked on my previous server, but I'm guessing my new server has a > newer version of php. > > What am I doing wrong? How can i pass the raw jpeg data to a function? What worked on your previous server? ImageJpeg has never returned the image data. It has always either written it to the file you specify, or if you don't specify one it will output it directly at that point. If you want to capture it in a variable you will need to use output buffering. See php.net/ob_start -Rasmus -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
