我可以让asyncio.sleep睡眠一年吗？

如何解决我可以让asyncio.sleep睡眠一年吗？

我想让asyncio.sleep()睡一年。

类似这样的东西：

async def run():
  # first print
  print('hi')
  # wait a year
  await asyncio.sleep(31536000)
  # second print
  print('hi after a year')

我显然不能等待一年来进行测试，但是理论上可以做到吗？

解决方法

您还可以通过编写单元测试并模拟协同程序来对此进行测试。

# file1.py
async def run(seconds):
  # first print
  print('hi')
  # wait a year
  await asyncio.sleep(seconds)
  # second print
  print('hi after a year')

# test_file1.py

import asyncio
from unittest import mock
from unittest.mock import patch

import pytest

import file1


class AsyncMock(mock.MagicMock):
    async def __call__(self,*args,**kwargs):
        return super(AsyncMock,self).__call__(*args,**kwargs)


class TestAsyncioSleep:
    @patch('file1.run',new_callable=AsyncMock)
    def test_fun(self,sleep_mock):
        event_loop = asyncio.get_event_loop()
        task = event_loop.run_until_complete(so_async.run(31536000))
        sleep_mock.assert_called_with(31536000)

我正在使用AsyncMock（）类嘲笑asyncio.sleep。