Django PIPE youtube-dl to view for download

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TL;DR: I want to pipe the output of youtube-dl to the user's browser on a button click, without having to save the video on my server's disk.

So I'm trying to have a "download" button on a page (django backend) where the user is able to download the video they're watching. I am using the latest version of youtube-dl. In my download view I have this piece of code:

    with youtube_dl.YoutubeDL(ydl_opts) as ydl:
        file = ydl.download([f"https://clips.twitch.tv/{pk}"])

And it works, to some extend. It does download the file to my machine, but I am not sure how to allow users to download the file. I thought of a few ways to achieve this, but the only one that really works for me would be a way to pipe the download to user(client) without needing to store any video on my disk. I found this issue on the same matter, but I am not sure how to make it work. I successfully piped the download to stdout using ydl_opts = {'outtmpl': '-'}, but I'm not sure how to pipe that to my view's response. One of the responses from a maintainer mentions a subprocess.Popen, I looked it up but couldn't make out how it should be implemented in my case.

youtube-dl django
1 Answers

I did a workaround. I download the file with a specific name, I return the view with HttpResponse, with force-download content-type, and then delete the file using python. It's not what I originally had in mind, but it's the second best solution that I could come up with. I will select this answer as accepted solution until a Python wizard gives a solution to the original question. The code that I have right now:

def download_clip(request, pk):
    ydl_opts = {
        'outtmpl': f"{pk}.mp4"
    }
    with youtube_dl.YoutubeDL(ydl_opts) as ydl:
        ydl.download([f"https://clips.twitch.tv/{pk}"])

    path = f"{pk}.mp4"
    file_path = os.path.join(path)
    if os.path.exists(file_path):
        with open(file_path, 'rb') as fh:
            response = HttpResponse(fh.read(), content_type="application/force-download")
            response['Content-Disposition'] = 'inline; filename=' + os.path.basename(file_path)
            os.remove(file_path)
            return response
    raise Http404
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