Re: passing a url to a page

Nothing happens no errors or anything I tried to debug using include('display_image.php?img_url=$image_url') I got a parse error. I would rather use buffers if this is possible? ----- Original Message ----- From: "Zoltán Németh" <znemeth@alterationx.hu> To: "Ross" <ross@aztechost.com> Cc: <php-general@lists.php.net> Sent: Friday, June 15, 2007 3:03 PM Subject: Re: [PHP] passing a url to a page 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?