I tried this code:
$image = new Imagick();
$f = fopen('http://www.url.com/image.jpg', 'rb');
$image = readImageFile($f);
but it doesn't work. Also, I believe it's not the best way to do it and I want the code to include best practices (now matter how ridiculous it sounds). On the same server, this code, using GD library, works just fine.
$im = imagecreatefromjpeg('http://www.url.com/image.jpg');
Did I miss something in Imagick manual?
try replacing:
$image = readImageFile($f)
with
$image->readImageFile($f);