Barryvdh\Snappy\ImageWrapper::stream PHP Метод

stream() публичный Метод

Return a response with the image to show in the browser
Устаревший: Use inline() instead
public stream ( string $filename = 'image.jpg' ) : Response
$filename string
Результат Symfony\Component\HttpFoundation\Response
    public function stream($filename = 'image.jpg')
    {
        return new StreamedResponse(function () {
            echo $this->output();
        }, 200, array('Content-Type' => 'image/jpeg', 'Content-Disposition' => 'inline; filename="' . $filename . '"'));
    }