download zip file and retrieve an office file that the name start with "source"

extract_zip_and_download_a_file.py

import io, requests, sys, zipfile
from wsgiref import simple_server


url_base = 'http://s3.amazonaws.com/xxxxxxxx'
s3_link = 'xxxxx.zip'
url = f"{url_base}/{s3_link}"


def get_content_type(filename):
    if filename.endswith('.xlsx'):
        return 'application/vnd.openxmlformats-officedocument.spreadsheetml.sheet'
    elif filename.endswith('.docx'):
        return 'application/vnd.openxmlformats-officedocument.wordprocessingml.document'
    elif filename.endswith('.pptx'):
        return 'application/vnd.openxmlformats-officedocument.presentationml.presentation'
    else:
        return 'text/plain'


def application(env, start_response):
    print(f'get {url}')
    r = requests.get(url)

    zipdata = io.BytesIO()
    zipdata.write(r.content)

    zfile = zipfile.ZipFile(zipdata)

    try:
        src_fn = [name for name in zfile.namelist() if name.startswith('source')].pop()
        content = zfile.open(src_fn)
    except:
        start_response('500 Internal Server Error', [('Content-type', 'text/plain')])
        return "source content wasn't found"

    start_response('200 OK', [('Content-type', get_content_type(src_fn))])
    return [content.read()]


if __name__ == '__main__':
    server = simple_server.make_server('', 8000, application)
    server.serve_forever()