You might also want to add a "quality" factor to the imagejpeg() function if the image "as is" looks a bit ratty. $factor="100" // or another suitable number between 100 and 0 imagejpeg($im, "test/im.jpg",$factor); hope this helps. Hugh
----- Original Message ----- From: "Andrew Chase" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Wednesday, July 31, 2002 1:10 PM Subject: RE: [PHP] Creating Image... > To write a GD image to disk as a jpeg, do this instead: > > > $im = imagecreate(50,100); > > imagejpeg($im, "test/im.jpg"); > > > You don't need to use file handlers. Of course, without doing anything like > adding text or lines to $im, it will just be written to disk as a blank > image. :) > > -Andy > > > -----Original Message----- > > From: mp [mailto:[EMAIL PROTECTED]] > > Sent: Wednesday, July 31, 2002 11:11 AM > > To: [EMAIL PROTECTED] > > Subject: [PHP] Creating Image... > > > > > > I want to create simple image... > > I execute this script: > > > > $im = imagecreate (50, 100); > > $fp = fopen("test/im.jpg","w"); > > fwrite($fp, imagegd($im)); > > fclose($fp); > > > > But there is some problems about this... > > Could somebody help me? > > Could somebody help me? > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
