remote URL .jpg

Christoph Grottolo wrote in message: > The manual says that you have to check if creation from url works. > Maybe you check first if your script works with a local jpg, then if > you can include a remote url (include > ("http://www.nytimes.com");). > To send the image you must use: > <? > header("Content-type: image/jpeg"); > imagejpeg($thenewimage,'',80); > imagedestroy($im); > ?> > You can not simply echo your image handler ($thenewimage), because > it's just a resource (pointer) and not a string. To use it you have to > convert it (back) into a common image format (jpeg, png...) > > Additionally you have to send a header in order to tell the browser > that you're going to send an image (and not a html page). > > Just read on in the manual, you know where... > > Christoph ............................................................... Christoph, I'm not sure what either the arguments that you supplied like: '' or 80 is in the function call below. ................................ imagejpeg($thenewimage,'',80); ............................... Please advise. Thank you. TR