asyncio在超时

时间:2016-10-27 22:01:50

标签: python python-3.4 python-asyncio

我有一个必须执行一些shell命令的脚本。但是,如果命令需要很长时间才能完成,则必须强行将其杀死。请考虑以下代码段:

import asyncio, random

q = asyncio.Queue()

MAX_WAIT = 5

@asyncio.coroutine
def blocking_task(sec):
    print('This task will sleep {} sec.'.format(sec))
    create = asyncio.create_subprocess_shell(
       'sleep {s}; echo "Woke up after {s} sec." >> ./tst.log'.format(s=sec),
        stdout=asyncio.subprocess.PIPE)
    proc = yield from create
    yield from proc.wait()

@asyncio.coroutine
def produce():
    while True:
        q.put_nowait(random.randint(3,8))
        yield from asyncio.sleep(0.5 + random.random())

@asyncio.coroutine
def consume():
    while True:
        value = yield from q.get()
        try:
            yield from asyncio.wait_for(blocking_task(value), MAX_WAIT)
        except asyncio.TimeoutError:
            print('~/~ Job has been cancelled !!')
        else:
            print('=/= Job has been done :]')


loop = asyncio.get_event_loop()
asyncio.ensure_future(produce())
asyncio.ensure_future(consume())
loop.run_forever()

此代码产生以下输出:

This task will sleep 4 sec.
=/= Job has been done :]
This task will sleep 8 sec.
~/~ Job has been cancelled !!
This task will sleep 5 sec.
~/~ Job has been cancelled !!

所以看起来它正如预期的那样工作,如果工作花了很长时间才能完成工作。但是,如果我检查日志,我可以确认,但是耗时的任务继续运行并且实际上没有停止/杀死/中止:

Woke up after 4 sec.
Woke up after 8 sec.
Woke up after 5 sec.

我希望日志中只有一行,因为其他进程必须在有机会完成之前中止:

Woke up after 4 sec.

有没有办法达到我想要的目的?

我甚至不确定我是否需要asyncio,也许concurrent.futures也可以使用。无论哪种方式任务都是相同的 - 终止任务,这需要花费太多时间来完成。

1 个答案:

答案 0 :(得分:1)

您可以使用Process.terminate

try:
    yield from proc.wait()
except asyncio.CancelledError:
    proc.terminate()
    raise

或者:

try:
    yield from proc.wait()
finally:
    if proc.returncode is None:
        proc.terminate()

修改

  

为什么我没有在代码中看到asyncio.CancelledError?

asyncio.wait_for(或其他任何内容)取消任务时,它会在相应的协程中抛出CancelledError。这允许协同程序在必要时执行一些清理(例如使用上下文管理器或try/finally子句)。不需要记录此错误,因为它是已取消任务的正常行为。但是在取消任务后尝试等待任务,并且会引发CancelledError