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

Question

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 securi

Answer 1

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