python:运行带有超时的进程并捕获stdout，stderr和退出状态 [英] python: run a process with timeout and capture stdout, stderr and exit status

问题描述

可能重复:
带有超时的子流程

Possible Duplicate:
subprocess with timeout

在Python中执行以下操作的最简单方法是什么:

What is the easiest way to do the following in Python:

• 运行外部进程
• 以字符串，stderr和退出状态捕获标准输出
• 设置超时时间.

我想要这样的东西:

import proc

try:
    status, stdout, stderr = proc.run(["ls", "-l"], timeout=10)
except proc.Timeout:
    print "failed"

推荐答案

我讨厌自己做这项工作.只需将其复制到您的proc.py模块中即可.

I hate doing the work by myself. Just copy this into your proc.py module.

import subprocess
import time
import sys

class Timeout(Exception):
    pass

def run(command, timeout=10):
    proc = subprocess.Popen(command, bufsize=0, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    poll_seconds = .250
    deadline = time.time()+timeout
    while time.time() < deadline and proc.poll() == None:
        time.sleep(poll_seconds)

    if proc.poll() == None:
        if float(sys.version[:3]) >= 2.6:
            proc.terminate()
        raise Timeout()

    stdout, stderr = proc.communicate()
    return stdout, stderr, proc.returncode

if __name__=="__main__":
    print run(["ls", "-l"])
    print run(["find", "/"], timeout=3) #should timeout

这篇关于python:运行带有超时的进程并捕获stdout，stderr和退出状态的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！