Question

我想使用超时来停止mqtt的阻止功能，我使用了timeout_decorator模块，它可以停止命令功能，但不能停止阻止功能subscribe.simple。

以下代码成功运行

import time
import timeout_decorator

@timeout_decorator.timeout(5, timeout_exception=StopIteration)
def mytest():
    print("Start")
    for i in range(1,10):
        time.sleep(1)
        print("{} seconds have passed".format(i))

if __name__ == '__main__':
    mytest()

结果如下：

Start
1 seconds have passed
2 seconds have passed
3 seconds have passed
4 seconds have passed
Traceback (most recent call last):
  File "timeutTest.py", line 12, in <module>
    mytest()
  File "/home/gyf/.local/lib/python3.5/site-packages/timeout_decorator/timeout_decorator.py", line 81, in new_function
    return function(*args, **kwargs)
  File "timeutTest.py", line 8, in mytest
    time.sleep(1)
  File "/home/gyf/.local/lib/python3.5/site-packages/timeout_decorator/timeout_decorator.py", line 72, in handler
    _raise_exception(timeout_exception, exception_message)
  File "/home/gyf/.local/lib/python3.5/site-packages/timeout_decorator/timeout_decorator.py", line 45, in _raise_exception
    raise exception()
timeout_decorator.timeout_decorator.TimeoutError: 'Timed Out'

但是我使用subscription.simple API失败

import timeout_decorator

@timeout_decorator.timeout(5)
def sub():
    # print(type(msg))
    print("----before simple")
    # threading.Timer(5,operateFail,args=)
    msg = subscribe.simple("paho/test/simple", hostname=MQTT_IP,port=MQTT_PORT,)
    print("----after simple")
    return msg


publish.single("paho/test/single", "cloud to device", qos=2, hostname=MQTT_IP,port=MQTT_PORT)
try:
    print("pub")
    msg = sub()
    print(msg)
except StopIteration as identifier:
    print("error")

结果无限等待

pub
----before simple

我希望包含subscription.simple API的功能可以在5秒钟后停止。

Answer 1

Asyncio 将无法在同一线程中处理阻塞功能。因此使用 asyncio.wait_for 失败。然而，受到 this blog post 的启发，我使用 loop.run_in_executor 来保持对阻塞线程的控制。

from paho.mqtt import subscribe
import asyncio

MQTT_IP = "localhost"
MQTT_PORT = 1883
msg = None


def possibly_blocking_function():
    global msg
    print("listenning for message")
    msg = subscribe.simple(
        "paho/test/simple",
        hostname=MQTT_IP,
        port=MQTT_PORT,
    )
    print("message received!")


async def main():
    print("----before simple")
    try:
        await asyncio.wait_for(
            loop.run_in_executor(None, possibly_blocking_function), timeout=5
        )
    except asyncio.TimeoutError:
        pass
    print("----after simple")

loop = asyncio.get_event_loop()
loop.run_until_complete(main())

输出：

----before simple
listenning for message
----after simple

请注意这并不完美，因为有正在运行的任务，程序不会结束。您可以使用各种解决方案退出它，但这超出了范围，因为我仍在寻找一种干净的方法来关闭卡住的线程。

如何使用超时停止阻止功能subscription.simple

1 个答案: