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Home/ Questions/Q 7595543
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T21:41:48+00:00 2026-05-30T21:41:48+00:00

I am using web.py to return a protocol buffer response from a post request

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I am using web.py to return a protocol buffer response from a post request and response time is critical. I have some writes to redis that I would like to do after the post response. rather than before.

r = redis.StrictRedis(host='localhost', port=6379, db=0)
class index:
    def POST(self):
    return pPbuffer
    r.set('a','b')

So, how can I modify the code so I can I can return as quickly as possible but doing post cleanup (no pun intended).

Thanks

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  1. Editorial Team
    Editorial Team
    2026-05-30T21:41:50+00:00Added an answer on May 30, 2026 at 9:41 pm

    If you are using wsgi or something as server you could use yield to generate contents time after time and the browser will receive them in sort.

    For your example:

    class index:
        def POST(self):
            yield pPbuffer
            r.set('a','b')
    

    And this is a good example which is doing it this way.

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