[PHP] Re: outputing image part 2
Thanks Stut we are getting somewhere.
Removing the header gives the full binary output of the file in IE and FF. I
also removes the echos and have the resized image showing in the browser. So
far so good. I know it works.
Now back to sending the URL.
I have this
$img_url=http://www.xxx.co.uk/images/ENbb24469/room1.JPG;;
echo img src=\common/display_image.php?img_url='$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.
Many thanks,
R.
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Re: [PHP] Re: outputing image part 2
Hi Ross,
Friday, June 15, 2007, 10:00:53 PM, you wrote:
I have this
$img_url=http://www.xxx.co.uk/images/ENbb24469/room1.JPG;;
echo img src=\common/display_image.php?img_url='$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
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http://www.corephp.co.uk
Never trust a computer you can't throw out of a window
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Re: [PHP] Re: outputing image part 2
Ross wrote:
Thanks Stut we are getting somewhere.
Removing the header gives the full binary output of the file in IE and FF. I
also removes the echos and have the resized image showing in the browser. So
far so good. I know it works.
Now back to sending the URL.
I have this
$img_url=http://www.xxx.co.uk/images/ENbb24469/room1.JPG;;
echo img src=\common/display_image.php?img_url='$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.
OK, your problem here is the single quotes in the img src around $img_url.
However, I question what you are actually doing. Is the image on the
same machine as this script? If so you should be using a filename not
the URL. Also, are you sure you're really saving anything by resizing
the image on your server like this? You should consider generating the
smaller version using a cron job and linking to it directly. Resizing
images on the fly is one of the quickest ways to destroy the scalability
of a website.
-Stut
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Re: [PHP] Re: outputing image part 2
On 6/15/07, Stut [EMAIL PROTECTED] wrote:
OK, your problem here is the single quotes in the img src around $img_url.
Stut's right. You should change:
echo img src=\common/display_image.php?img_url='$img_url'\
width=\200\ height=\100\ /;
to:
echo img
src=\common/display_image.php?img_url=.$img_url.\ width=\200\
height=\100\ /;
He's also right about the dynamic resizing of images causing a
serious strain on the server. Each time it has to do that, it uses up
a significant amount of resources for a seemingly quick and small
operation. Multiply that by n number of simultaneous accesses and you
can see the exponential - and potentially devastating - pitfallic
results.
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[mobile] (570-) 766-8107
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Re: [PHP] Re: outputing image part 2
I am taking your advice.
I am going to rezize any image greater than 400px wide prior to upload. I
have to combine my upload script
for ($i=0; $i $total; $i++) {
$fileName = $_FILES['userfile']['name'][$i];
$tmpName = $_FILES['userfile']['tmp_name'][$i];
$fileSize = $_FILES['userfile']['size'][$i];
$fileType = $_FILES['userfile']['type'][$i];
$position=$i+1;
$img_url = images/$customer_id/. basename(
$_FILES['userfile']['name'][$i]);
if(move_uploaded_file($_FILES['userfile']['tmp_name'][$i], $img_url)) {
chmod('images/'.$customer_id, 0777);
chmod($img_url, 0777);
}
with my resize script
$image = imagecreatefromjpeg($img_url);
if ($image === false) { exit; }
// Get original width and height
$width = imagesx($image);
$height = imagesy($image);
// New width and height
$new_width = 200;
$new_height = 150;
// Resample
$image_resized = imagecreatetruecolor($new_width, $new_height);
imagecopyresampled($image_resized, $image, 0, 0, 0, 0, $new_width,
$new_height, $width, $height);
Any ideas how to save the imagecopyresampled() to the folder?
T
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Re[2]: [PHP] Re: outputing image part 2
Hi Ross, Friday, June 15, 2007, 10:40:37 PM, you wrote: Any ideas how to save the imagecopyresampled() to the folder? Call imagepng (or imagejpeg or whatever) and pass it a filename to save the image instead of output it. Check the help files for examples. Cheers, Rich -- Zend Certified Engineer http://www.corephp.co.uk Never trust a computer you can't throw out of a window -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
