Return file to WSGI GET request

Question

I am new to WSGI in python; but there is a windows server on which isapi_wsgi is installed. I also have a script that processes my GET requests and works fine. The fact is that someone sends me a request, and I need to return the zip file to the requestor. The following code is in my GET handler, and it works, but does not seem to be the correct way to return a zip file:

  # open zip file return it
  fin = open(zOutFilename, "rb")
  start_response( "200 OK", [('Content-Type', 'application/zip')])
  return fin.read()

The fact is that you are returning a "stream" - this means that you are losing the file name (the browser simply calls it the GET request name), and it seems to be very slow.

Is there a better way to return the file to be downloaded using wsgi, then this method?

Thanks in advance

Answer 1

Taken directly from PEP 333 :

if 'wsgi.file_wrapper' in environ:
    return environ['wsgi.file_wrapper'](filelike, block_size)
else:
    return iter(lambda: filelike.read(block_size), '')

Also, you probably need a Content-Disposition header to provide the file name to the client.