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How could I use requests in asyncio?

I want to do parallel http request tasks in asyncio, but I find that python-requests would block the event loop of asyncio. I've found aiohttp but it couldn't provide the service of http request using a http proxy.

So I want to know if there's a way to do asynchronous http requests with the help of asyncio.

like image 405
flyer Avatar asked Oct 17 '22 08:10

flyer


2 Answers

To use requests (or any other blocking libraries) with asyncio, you can use BaseEventLoop.run_in_executor to run a function in another thread and yield from it to get the result. For example:

import asyncio
import requests

@asyncio.coroutine
def main():
    loop = asyncio.get_event_loop()
    future1 = loop.run_in_executor(None, requests.get, 'http://www.google.com')
    future2 = loop.run_in_executor(None, requests.get, 'http://www.google.co.uk')
    response1 = yield from future1
    response2 = yield from future2
    print(response1.text)
    print(response2.text)

loop = asyncio.get_event_loop()
loop.run_until_complete(main())

This will get both responses in parallel.

With python 3.5 you can use the new await/async syntax:

import asyncio
import requests

async def main():
    loop = asyncio.get_event_loop()
    future1 = loop.run_in_executor(None, requests.get, 'http://www.google.com')
    future2 = loop.run_in_executor(None, requests.get, 'http://www.google.co.uk')
    response1 = await future1
    response2 = await future2
    print(response1.text)
    print(response2.text)

loop = asyncio.get_event_loop()
loop.run_until_complete(main())

See PEP0492 for more.

like image 228
christian Avatar answered Oct 18 '22 22:10

christian


aiohttp can be used with HTTP proxy already:

import asyncio
import aiohttp


@asyncio.coroutine
def do_request():
    proxy_url = 'http://localhost:8118'  # your proxy address
    response = yield from aiohttp.request(
        'GET', 'http://google.com',
        proxy=proxy_url,
    )
    return response

loop = asyncio.get_event_loop()
loop.run_until_complete(do_request())
like image 106
mindmaster Avatar answered Oct 18 '22 21:10

mindmaster