flask_dynamic_file_serve.py

import io
import tarfile
import os

import requests
from flask import Flask, request, send_file

app = Flask(__name__)

def create_tarfile(urls):
    tar_stream = io.BytesIO()
    with tarfile.open(mode='w', fileobj=tar_stream) as tar:
        for url in urls:
            r = requests.get(url)
            with io.BytesIO(r.content) as f:
                name = os.path.basename(url)
                info = tarfile.TarInfo(name)
                info.size = len(r.content)
                tar.addfile(info, f)
    tar_stream.seek(0)
    return tar_stream

@app.route('/')
def index():
    urls = request.args.getlist('url')
    f = create_tarfile(urls) # f is intentionally not closed. Would flask close this?
    return send_file(f, attachment_filename='foobar.tar')

if __name__ == '__main__':
    app.run(port=5001, debug=True)