我可以在不启用fopen包装器的情况下将URL用作imagecreatefromjpeg()的源吗? [英] Can I use a URL as the source for imagecreatefromjpeg() without enabling fopen wrappers?

问题描述

我知道可以使用带有URL的imagecreatefromjpeg()，imagecreatefrompng()等作为fopen()的文件名"，但是由于安全问题，我无法启用包装器.有没有办法在不启用URL的情况下将URL传递给imagecreatefromX()?

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?

我也尝试过使用cURL，这也给我带来了问题:

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);

推荐答案

您可以使用cURL下载文件，然后将结果通过管道传输到imagecreatefromstring.

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

示例:

    $ch = curl_init();
    curl_setopt($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);

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