Flask send stream as response

问题

I'm trying to "proxy" my Flask server (i will call it Server#01) with another server(Server#02). It's working well except for one thing : when the Server#01 use send_from_directory(), i don't know how to re-send this file.

My classic "proxy"

result = requests.get(my_path_to_server01)
return Response(stream_with_context(result.iter_content()), 
                content_type = result.headers['Content-Type'])

With a file a response, it's taking hours... So i tried many things. The one who work is :

result = requests.get(my_path_to_server01, stream=True)

with open('img.png', 'wb') as out_file:
    shutil.copyfileobj(result.raw, out_file)

return send_from_directory('./', 'img.png')

I would like to "redirect" my response ("result" variable), or send/copy a stream of my file. Anyways I don't want to use a physical file because it don't seems the proper way in my mind and i can imagine all problems who can happens because of that.

回答1:

There should not be any problem with your "classic" proxy other than that it should use stream=True, and specify a chunk_size for response.iter_content().

By default chunk_size is 1 byte, so the streaming will be very inefficient and consequently very slow. Trying a larger chunk size, e.g. 10K should yield faster transfers. Here's some code for the proxy.

import requests
from flask import Flask, Response, stream_with_context

app = Flask(__name__)

my_path_to_server01 = 'http://localhost:5000/'

@app.route("/")
def streamed_proxy():
    r = requests.get(my_path_to_server01, stream=True)
    return Response(r.iter_content(chunk_size=10*1024),
                    content_type=r.headers['Content-Type'])

if __name__ == "__main__":
    app.run(port=1234)

You don't even need to use stream_with_context() here because you don't need access to the request context within the generator returned by iter_content().

来源：https://stackoverflow.com/questions/39272072/flask-send-stream-as-response