Re: Re: outputing image part 2

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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"

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