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.