Well, my old code isn't working with my new install of PHP 4.3.2 and Apache2. The GD php_gd2.dll is loaded... however even this line of code wont display a jpeg file:
<?
header("Content-type: image/jpeg"
;
$im = imagecreatefromjpeg("bg1.jpg"
;
imagejpeg($im);
?>
The bg1.jpg is in the same directory as the script...
Any ideas?
- "Delightfully confusing..." raves The New York Times
-kas
<?
header("Content-type: image/jpeg"
$im = imagecreatefromjpeg("bg1.jpg"
imagejpeg($im);
?>
The bg1.jpg is in the same directory as the script...
Any ideas?
- "Delightfully confusing..." raves The New York Times
-kas