Unzip or inflate php://input stream?

Question

I'm trying to unzip a zip file directly from the php://input stream. I'm running Laravel Homestead, PHP 7.1.3-3+deb.sury.org~xenial+1, with an endpoint at myproject.app/upload, here is the curl command:

curl --request POST \
  --url 'http://myproject.app/upload' \
  --data-binary "@myfile.zip" \

Here is a list of all the methods I've tried, which all fail:

dd(file_get_contents('compress.zlib://php://input'));

file_get_contents(): cannot represent a stream of type Input as a File Descriptor

$fh = fopen('php://input', 'rb');

stream_filter_append($fh, 'zlib.inflate', STREAM_FILTER_READ, array('window'=>15));

$data = '';

while (!feof($fh)) {
    $data .= fread($fh, 8192);
}

dd($data);

""

$zip = new ZipArchive;

$zip->open('php://input');
$zip->extractTo(storage_path() . '/' . 'myfile');
$zip->close();

ZipArchive::extractTo(): Invalid or uninitialized Zip object

Here are all the links I've found on the subject:

http://php.net/manual/en/wrappers.php#83220

http://php.net/manual/en/wrappers.php#109657

http://php.net/manual/en/wrappers.compression.php#118461

https://secure.phabricator.com/rP42566379dc3c4fd01a73067215da4a7ca18f9c17

https://arjunphp.com/how-to-unpack-a-zip-file-using-php/

I'm beginning to think that it's not possible to operate on streams with PHP's built-in zip functionality. The overhead and complexity of writing temporary files would be pretty disappointing. Does anyone know how to do this, or is it a bug?

Unzip or inflate php://input stream?

1 Answers