如何使用当前的 php 页面 URL 作为外部图像的参考？

Question

I'm trying to download an external image on my server with the variable that is written in the url, for example:

www.myserver.com/script.php/imageurl="https://www.otherserver.com/image.png"

for this I am trying to use the following PHP code:

1: header('Content-type: image/png');
2: $imageurl = $_GET['imageurl'];
3: $remote_image = file_get_contents($imageurl);
4: file_put_contents("/tmp/result.png", $remote_image);

The problem is obvious, the page returns an error because it can not find anything in the path specified in the url...

It's possible to accomplish this and "ignore" the url that comes after script.php/?

Answer 1

First change the link from

www.myserver.com/script.php/imageurl="www.otherserver.com/image.png"

TO

www.myserver.com/script.php?imageurl="www.otherserver.com/image.png"

You can download an image like so:

SOURCE

If you have allow_url_fopen set to true:

$url = 'http://example.com/image.php';
$img = '/my/folder/flower.gif';
file_put_contents($img, file_get_contents($url));

OR

copy('http://example.com/image.php', 'local/folder/flower.jpg');

Else use cURL :

$ch = curl_init('http://example.com/image.php');
$fp = fopen('/my/folder/flower.gif', 'wb');
curl_setopt($ch, CURLOPT_FILE, $fp);
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_exec($ch);
curl_close($ch);
fclose($fp);