Question

Just wondering if there is any difference with the two below code examples:

$image = 'http://www.example.com/image.jpg'

$photo = file_get_contents($image);

ob_start();
header("Content-type: image/jpeg");
print($photo);
ob_end_flush(); 

or…

$image = 'http://www.example.com/image.jpg'

$photo = file_get_contents($image);

ob_start();
header("Content-type: image/jpeg");
readfile($photo);
ob_end_flush();

Answer 1

readfile‘s parameter is a filename, not the content itself. Therefore, you’d call it like this:

$image = 'http://www.example.com/image.jpg'
ob_start();
header("Content-type: image/jpeg");
readfile($image);
ob_end_flush();

Since readfile reads and write chunks at a time, its memory consumption will be constant, whereas when you store the result of file_get_contents into $photo, you’ll need to have enough memory to store the image.

In your case, the output buffering makes the file_get_contents variant demand twice as much memory as the image’s size. For a large image, readfile therefore halves the memory requirements. Note that your output buffering means that the download will be delayed. If you don’t need it for anything else, you will get better performance (both in actual speed and server memory requirements) if you simply disable it:

$image = 'http://www.example.com/image.jpg'
header("Content-type: image/jpeg");
if (@readfile($image) === false) {
   // print error image
}