尝试使用 run_forever 创建一个永远运行的异步函数

Question

I am trying to access a websocket to listen to some information and so I want to use the run_forever command to keep listening until I stop the program, but when I do I get the error below, could someone tell me what I am doing wrong?

Code:

loop = asyncio.get_event_loop()

async def listen():
    url = "websocket url starting with wss"

    async with websockets.connect(url) as ws:
        msg = await ws.recv()
        print(msg)


loop.run_forever(listen())

the error then says:

RuntimeWarning: coroutine 'listen' was never awaited
  loop.run_forever(listen())
RuntimeWarning: Enable tracemalloc to get the object allocation traceback
Traceback (most recent call last):
  File "C:\Users\...", line 18, in <module>
    loop.run_forever(listen())
TypeError: run_forever() takes 1 positional argument but 2 were given

Answer 1

I tried loading your script into Python 3.9.6 and i saw your error.

This is how i got it to run (you need to enter the proper URL):

import asyncio, websockets

async def listen():
    url = "websocket url starting with wss"
    async with websockets.connect(url) as ws:
        msg = await ws.recv()
        print(msg)

async def main():
    loop = asyncio.get_event_loop()
    loop.run_forever(await listen())

if __name__ == "__main__":
    asyncio.run(main())

I hope it can help you to get further with your program.