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@xxxxxxxxxxxxxx>
To: "Ross" <ross@xxxxxxxxxxxxx>
Cc: <php-general@xxxxxxxxxxxxx>
Sent: Friday, June 15, 2007 3:03 PM
Subject: Re: 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?
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