2007. 06. 15, péntek keltezéssel 14.45-kor Ross ezt írta: > I have a display_image.php page > > <?php > > > > $image = imagecreatefromjpeg($img_url); where do you get $img_url from? GET request? then do something like $img_url = $_GET['img_url']; if (!file_exists($img_url)) die "bad hacker"; before the above line. this will let you get rid of register_globals, and does a basic check of input, which is always very important > if ($image === false) { exit; } > > // Get original width and height > echo $width = imagesx($image); > echo $height = imagesy($image); > > // New width and height > $new_width = 200; > $new_height = 150; > > // Resample > $image_resized = imagecreatetruecolor($new_width, $new_height); > imagecopyresampled($image_resized, $image, 0, 0, 0, 0, $new_width, > $new_height, $width, $height); > > // Display resized image > ; > header('Content-type: image/jpeg'); > imagejpeg($image_resized); > exit(); > > ?> > > > I want to output this as an image but cannot get it working. I need to pass > the image url something like this I thought would work > > echo "<img src=\"display_image.php?img_url=$image_url>"; this should work. what do you mean by not working? do you get any error message? anything? without information no one can tell you what's wrong... greets Zoltán Németh > > > any ideas? > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php