如何将多个值从异步函数返回到同步上下文？

Question

我需要从异步调用中收集结果并将它们返回给一个普通函数——桥接异步和同步代码部分让我感到困惑。

首先尝试，使用丑陋的 inout 参数。

import asyncio
import aiohttp

async def one_call(url):
    async with aiohttp.ClientSession() as session:
        async with session.get(url) as response:
            txt = await response.text()
            return txt[0:20]


async def do_all(result_inout):
    urls = ["https://cnn.com", "https://nyt.com", "http://reuters.com"]
    out = await asyncio.gather(*[one_call(url) for url in urls])
    result_inout += out

if __name__ == "__main__":
    result_inout = []
    asyncio.run(do_all(result_inout))

    print(result_inout)

第二次尝试，但直接使用事件循环，不鼓励应用程序代码。

if __name__ == "__main__":
    # same imports, same one_call, same urls
    loop = asyncio.get_event_loop()

    aggregate_future = asyncio.gather(*[one_call(url) for url in urls])
    results = loop.run_until_complete(aggregate_future)
    loop.close()

    print(results) 

做这个的最好方式是什么？

Answer 1

不需要result_inout，你可以只使用out = asyncio.run(do_all())来获得return res的do_all。

import asyncio
import aiohttp

async def one_call(url):
    await asyncio.sleep(2)
    return 0


async def do_all():
    urls = ["https://cnn.com", "https://nyt.com", "http://reuters.com"]
    out = await asyncio.gather(*[one_call(url) for url in urls])
    return out

if __name__ == "__main__":
    out = asyncio.run(do_all())

    print(out)

出来会[0, 0, 0]。