如何使用 python 2.7.6 使 subprocess.call 超时?

A simple way I've always done timeouts with 2.7 is utilizing subprocess.poll() alongside time.sleep() with a delay. Here's a very basic example:

import subprocess
import time

x = #some amount of seconds
delay = 1.0
timeout = int(x / delay)

args = #a string or array of arguments
task = subprocess.Popen(args)

#while the process is still executing and we haven't timed-out yet
while task.poll() is None and timeout > 0:
     #do other things too if necessary e.g. print, check resources, etc.
     time.sleep(delay)
     timeout -= delay

如果您设置 x = 600，那么您的超时将达到 10 分钟.而 task.poll() 会查询进程是否已经终止.time.sleep(delay) 在这种情况下会休眠 1 秒，然后将超时时间减 1 秒.您可以随心所欲地玩弄这部分，但基本概念始终相同.