Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
599 views
in Technique[技术] by (71.8m points)

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

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);
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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);

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...