尝试使用 run_forever 创建一个永远运行的异步函数

Question

我正在尝试访问 websocket 来监听一些信息，所以我想使用 run_forever 命令继续监听直到我停止程序，但是当我收到下面的错误时，有人能告诉我我做错了什么吗？

代码：

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())

然后错误说：

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

我尝试将您的脚本加载到 Python 3.9.6 中，但我看到了您的错误。

这是我运行它的方式（您需要输入正确的 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())

我希望它可以帮助您进一步了解您的程序。