Re: Re: outputing image part 2

Hi Ross, Friday, June 15, 2007, 10:00:53 PM, you wrote: > I have this > > $img_url="http://www.xxxxxxx.co.uk/images/ENbb24469/room1.JPG"; > echo "<img src=\"common/display_image.php?img_url='$img_url'\" > width=\"200\" height=\"100\" />>"; > and on the display image page I have: > $img_url= $_GET['img_url']; > $image = imagecreatefromjpeg($img_url); > if ($image === false) { exit; } > /// rest of the resize code goes here..... > This still does not work but if I plug the url in manually at the top of > display_image.php it works, this suggests it is not being sent properly. You are sending a complete URL to your image, not a path. This means that if your installation of PHP does NOT have the ability to open files from URLs (allow_url_fopen) then it won't be able to open the image from the location you gave it. If you want to load in images from other web sites then make sure allow_url_fopen is enabled in PHP. If you don't want to do this, then don't pass in a URL, pass in the *path* to the image instead. Cheers, Rich -- Zend Certified Engineer http://www.corephp.co.uk "Never trust a computer you can't throw out of a window"