How to load an image resource from a URL into a PHP script, using cURL?
Solution 1
What you need instead of imagecreatefrompng()
is imagecreatefromstring()
, because the former expects a filename instead of the file contents itself.
Solution 2
This worked for me
$image = imagecreatefromstring(file_get_contents('http://chart.googleapis.com/chart?cht=qr&chs=500x500&chl=xghsdfgsdfg&choe=UTF-8&chld=L|0'));
header('Content-Type: image/png');
imagepng($image);
Note: I had to use http rather than https because I haven't set up ssl on my local server.
Related videos on Youtube
Andrei Oniga
Striving to become one of the best, one step at a time.
Updated on June 04, 2022Comments
-
Andrei Oniga almost 2 years
I'm using the Google Charts service to generate some QR codes that I afterward need to manipulate (e.g. rotate, scale) in a PHP script and merge with other images to generate one final image.
How do I correctly load such a resource (from a URL) into a PHP script, in a way that will allow me to manipulate it?
An example URL is: https://chart.googleapis.com/chart?cht=qr&chs=500x500&chl=xghsdfgsdfg&choe=UTF-8&chld=L|0
I currently have the following code to retrieve the image using cURL:
function getImage($url){ $ch = curl_init ($url); curl_setopt($ch, CURLOPT_HEADER, 0); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch, CURLOPT_BINARYTRANSFER,1); $resource = curl_exec($ch); curl_close ($ch); return $resource; }
But when I use it like this:
$image = imagecreatefrompng(getImage("https://chart.googleapis.com/chart?cht=qr&chs=500x500&chl=xghsdfgsdfg&choe=UTF-8&chld=L|0"));
The following error is returned:
Warning: imagecreatefrompng(‰PNG ) [function.imagecreatefrompng]: failed to open stream: No such file or directory in /home/picselbc/public_html/projects/cakemyface/preview.php on line 383 https://chart.googleapis.com/chart?cht=qr&chs=500x500&chl=xghsdfgsdfg&choe=UTF-8&chld=L|0
-
Andrei Oniga over 11 yearsYou're right, that works! Any idea what I can do if I use the SimpleImage class though? (link). Because I'd need to use it like so:
$image = new SimpleImage(getImage($url));
. -
Ja͢ck over 11 yearsIf
SimpleImage
needs a file you have to write one in a temporary location perhaps? -
Andrei Oniga over 11 yearsI was afraid you would say that :)