How do I stream a file using werkzeug?

python werkzeug
Frankie Ribery picture Frankie Ribery · Mar 2, 2011 · Viewed 8.9k times · Source

I want to stream a big file via werkzeug.
Currently my wsgi application looks like this:

from werkzeug.wrappers import Request, Response
from werkzeug.wsgi import ClosingIterator, wrap_file
import os

class Streamer(object):

    def __init__(self):
        pass

    def __call__(self, environ, start_response):
        request = Request(environ)
        filename = os.getcwd() + "/bigfile.xml"
        try:
            response = wrap_file(environ, open(filename) )
            return response
        except HTTPException, e:
            response = e
            return ClosingIterator(response(environ, start_response))

I'm not sure what I should do with the object returned by the wrap_file function.

Answer

Shekhar picture Shekhar · Mar 2, 2011

Haven't tried myself but I think following will work.

g = file(path_to_bigfile) # or any generator
return Response(g, direct_passthrough=True)

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