如何在没有等待的情况下调用异步功能？

Question

我在aiohttp应用程序中有一个控制器动作.

async def handler_message(request):

    try:
        content = await request.json()
        perform_message(x,y,z)
    except (RuntimeError):
        print("error in perform fb message")
    finally:
        return web.Response(text="Done")

perform_message是异步函数.现在,当我调用动作时,我希望我的动作尽快返回,并且perform_message放入事件循环.

这样,不执行perform_message

Answer 1

最简单的形式：

import asyncio
from datetime import datetime

def _log(msg : str):
    print(f"{datetime.utcnow()} {msg}")

async def dummy(name, delay_sec):
    _log(f"{name} entering ...")
    await asyncio.sleep(delay_sec)
    _log(f"{name} done for the day!")

async def main():
    asyncio.create_task(dummy('dummy1', 5)) # last to finish
    asyncio.create_task(dummy('dummy2', 3)) # 2nd
    asyncio.create_task(dummy('dummy3', 1)) # First to finish

    _log(f"Yo I didn't wait for ya!")
    await asyncio.sleep(10)

asyncio.get_event_loop().run_until_complete(main())

输出：

2022-09-18 00:53:13.428285 Yo I didn't wait for ya!
2022-09-18 00:53:13.428285 dummy1 entering ...
2022-09-18 00:53:13.428285 dummy2 entering ...
2022-09-18 00:53:13.428285 dummy3 entering ...
2022-09-18 00:53:14.436801 dummy3 done for the day!
2022-09-18 00:53:16.437226 dummy2 done for the day!
2022-09-18 00:53:18.424755 dummy1 done for the day!

Answer 2

一种方法是使用create_task函数:

import asyncio

async def handler_message(request):
    ...
    loop = asyncio.get_event_loop()
    loop.create_task(perform_message(x,y,z))
    ...