Maybe this helps:
$file = "/tmp/pathtoimage/image.jg"
$fp = fopen($file, filesize($file));
$image = fread($fp, filesize($file)); //not sure about fread function¿
print $image //or imagejpeg($image), I'm not sure..
Grtz,
Luc
[EMAIL PROTECTED]
- Original Message -
From: "Mark Gordon" <[
No no no, what you want to do is echo $row[0], not .
$row[0] is the actual jpeg data, and the browser will interpret it like that
since you put the header("Content-type: text/jpeg");
Try echoing $row[0] only and tell me how that goes.
-- David
-Original Message-
From: Mark Gordon [mailt
