Can I use a URL as the source for imagecreatefromjpeg() without enabling fopen wrappers?

问题

I know it’s possible to use imagecreatefromjpeg(), imagecreatefrompng(), etc. with a URL as the ‘filename’ with fopen(), but I'm unable to enable the wrappers due to security issues. Is there a way to pass a URL to imagecreatefromX() without enabling them?

I’ve also tried using cURL, and that too is giving me problems:

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL,"http://www.../image31.jpg"); //Actually complete URL to image
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$data = curl_exec($ch);
curl_close($ch);

$image = imagecreatefromstring($data);
var_dump($image);

imagepng($image);
imagedestroy($image);

回答1:

You can download the file using cURL then pipe the result into imagecreatefromstring.

Example:

    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $imageurl); 
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); // good edit, thanks!
    curl_setopt($ch, CURLOPT_BINARYTRANSFER, 1); // also, this seems wise considering output is image.
    $data = curl_exec($ch);
    curl_close($ch);

    $image = imagecreatefromstring($data);

回答2:

You could even implement a cURL based stream wrapper for 'http' using stream_wrapper_register.

回答3:

You could always download the image (e.g. with cURL) to a temporary file, and then load the image from that file.

来源：https://stackoverflow.com/questions/12337639/can-i-use-a-url-as-the-source-for-imagecreatefromjpeg-without-enabling-fopen-w