I have manually put in the url my display_image.php page to debug as
sugested and all I get is the URL of the display_image.php page output on
the screen.
This is what I see
http://xxx.co.uk/common/display_image.php
this is the display_image.php file:
$img_url="http://www.xxx
Zoltán Németh wrote:
> 2007. 06. 15, péntek keltezéssel 14.45-kor Ross ezt írta:
>> I have a display_image.php page
>>
>> >
>>
>>
>> $image = imagecreatefromjpeg($img_url);
>
> where do you get $img_url from? GET request?
> then do something like
>
> $img_url = $_GET['img_url'];
> if (!file_exist
Daniel Brown wrote:
$image_url = 'http://www.pr0nsite.com/gallery/big-boobs.jpg';
Well that was just too disappointing for a Friday afternoon. Ya tease!
-Stut
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nal Message -
> From: "Zoltán Németh" <[EMAIL PROTECTED]>
> To: "Ross" <[EMAIL PROTECTED]>
> Cc:
> Sent: Friday, June 15, 2007 3:03 PM
> Subject: Re: [PHP] passing a url to a page
>
>
> 2007. 06. 15, péntek keltezéssel 14.45-kor
On 6/15/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
Nothing happens no errors or anything I tried to debug using
include('display_image.php?img_url=$image_url') I got a parse error.
You can't include a file with a query string attached. However,
you /can/ do:
http://www.pr0nsite.com/g
[EMAIL PROTECTED] wrote:
Nothing happens no errors or anything I tried to debug using
include('display_image.php?img_url=$image_url') I got a parse error.
I would rather use buffers if this is possible?
You can't parse GET parameters with include. Try this...
$_GET['img_url'] = $image_url;
i
" <[EMAIL PROTECTED]>
Cc:
Sent: Friday, June 15, 2007 3:03 PM
Subject: Re: [PHP] passing a url to a page
2007. 06. 15, péntek keltezéssel 14.45-kor Ross ezt írta:
I have a display_image.php page
where do you get $img_url from? GET request?
then do something like
$img_url = $_GET[&
I have a display_image.php page
I want to output this as an image but cannot get it working. I need to pass
the image url something like this I thought would work
echo "";
any ideas?
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2007. 06. 15, péntek keltezéssel 14.45-kor Ross ezt írta:
> I have a display_image.php page
>
>
>
>
> $image = imagecreatefromjpeg($img_url);
where do you get $img_url from? GET request?
then do something like
$img_url = $_GET['img_url'];
if (!file_exists($img_url)) die "bad hacker";
before
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