note 55987 added to function.imagejpeg

If you wish to capture the jpg data into a variable, rather than outputting it or saving it into a file (perhaps so you can put it in a database), you might want to consider output buffering. Something along these lines should work: <?php ob_start(); // start a new output buffer imagejpeg( $newimage, NULL, 100 ); $ImageData = ob_get_contents(); ob_end_clean; // stop this output buffer ?> ---- Manual Page -- http://www.php.net/manual/en/function.imagejpeg.php Edit -- http://master.php.net/manage/user-notes.php?action=edit+55987 Delete: added to the manual -- http://master.php.net/manage/user-notes.php?action=delete+55987&report=yes&reason=added+to+the+manual Delete: bad code -- http://master.php.net/manage/user-notes.php?action=delete+55987&report=yes&reason=bad+code Delete: spam -- http://master.php.net/manage/user-notes.php?action=delete+55987&report=yes&reason=spam Delete: useless -- http://master.php.net/manage/user-notes.php?action=delete+55987&report=yes&reason=useless Delete: non-english -- http://master.php.net/manage/user-notes.php?action=delete+55987&report=yes&reason=non-english Delete: other reasons -- http://master.php.net/manage/user-notes.php?action=delete+55987&report=yes Reject -- http://master.php.net/manage/user-notes.php?action=reject+55987&report=yes Search -- http://master.php.net/manage/user-notes.php