I have a display_image.php page <?php
$image = imagecreatefromjpeg($img_url); 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>"; any ideas? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php