php://input as resouce AND string - php

Question

Whit the follow code:
$file = fopen('php://input', 'r');
$temp = tmpfile();
$imageSize = stream_copy_to_stream($file, $temp);
$imageDimensions = getimagesize($file); // Error here
$imageInfos = pathinfo($_GET['selected-image']);
I get this error
getimagesize() expects parameter 1 to be string, resource given
Because $file is a resource of fopen. How I can have a resource and a string if I can't read php://input twice?
UPDATE
I have tried this:
$file = fopen('php://input', 'r');
$tempName = tempnam(sys_get_temp_dir(), '.upload');
$imageSize = fwrite(fopen($tempName, 'w+'), stream_get_contents($file));
$imageDimensions = getimagesize($tempName);
$imageInfos = pathinfo($_GET['selected-image']);
// Unlinks and other stuffs
The error:
Notice: getimagesize() [function.getimagesize]: Read error! in
Thanks!

Answer 1

Because getimagesize() only accepts string filenames, not resources, you will need to use tempnam() to create a temporary file with a name that you can pass to getimagesize().
$file = fopen('php://input', 'r');
$tempName = tempnam('/var/tmp', 'img_');
$temp = fopen($tempName, 'w');
$imageSize = stream_copy_to_stream($file, $temp);
fclose($temp);
$imageDimensions = getimagesize($tempName);
This assumes that the user running PHP has write and read permissions to the directory /var/tmp, but if it doesn't you can change to a directory that the user does have r/w permissions to.

Answer 2

Use tempnam() to generate your temp file, instead of tmpfile(). Then you have a file path which you can pass to getimagesize():
$file = fopen('php://input', 'r');
$tempname = tempnam('.', 'img');
$tempfile = fopen($tempname, 'w+');
$imageSize = stream_copy_to_stream($file, $tempfile);
$imageDimensions = getimagesize($tempname);
$imageInfos = pathinfo($_GET['selected-image']);
// At the end of the script, you need to remember to...
fclose($tempfile);
unlink($tempname);