下面是运行任意命令返回其标准输出数据的Python代码,或在非零退出码上引发异常:

proc = subprocess.Popen(
    cmd,
    stderr=subprocess.STDOUT,  # Merge stdout and stderr
    stdout=subprocess.PIPE,
    shell=True)

communication用于等待进程退出:

stdoutdata, stderrdata = proc.communicate()

子进程模块不支持超时——杀死运行超过X秒的进程的能力——因此,通信可能需要很长时间才能运行。

在Windows和Linux上运行的Python程序中实现超时的最简单方法是什么?


当前回答

对于python 2.6+,使用gevent

 from gevent.subprocess import Popen, PIPE, STDOUT

 def call_sys(cmd, timeout):
      p= Popen(cmd, shell=True, stdout=PIPE)
      output, _ = p.communicate(timeout=timeout)
      assert p.returncode == 0, p. returncode
      return output

 call_sys('./t.sh', 2)

 # t.sh example
 sleep 5
 echo done
 exit 1

其他回答

timeout现在由子进程模块中的call()和communication()支持(从Python3.3开始):

import subprocess

subprocess.call("command", timeout=20, shell=True)

这将调用该命令并引发异常

subprocess.TimeoutExpired

如果命令在20秒后还没有完成。

然后你可以处理异常来继续你的代码,就像这样:

try:
    subprocess.call("command", timeout=20, shell=True)
except subprocess.TimeoutExpired:
    # insert code here

希望这能有所帮助。

虽然我还没有广泛地研究它,但我在ActiveState中发现的这个装饰器似乎对这类事情非常有用。伴随着subprocess.Popen(…, close_fds=True),至少我已经准备好在Python中编写shell脚本。

我有一个问题,我想终止一个多线程子进程,如果它花费的时间超过给定的超时长度。我想在Popen()中设置一个超时,但它不起作用。然后,我意识到Popen().wait()等于call(),所以我有了在.wait(timeout=xxx)方法中设置超时的想法,这最终工作了。因此,我是这样解决的:

import os
import sys
import signal
import subprocess
from multiprocessing import Pool

cores_for_parallelization = 4
timeout_time = 15  # seconds

def main():
    jobs = [...YOUR_JOB_LIST...]
    with Pool(cores_for_parallelization) as p:
        p.map(run_parallel_jobs, jobs)

def run_parallel_jobs(args):
    # Define the arguments including the paths
    initial_terminal_command = 'C:\\Python34\\python.exe'  # Python executable
    function_to_start = 'C:\\temp\\xyz.py'  # The multithreading script
    final_list = [initial_terminal_command, function_to_start]
    final_list.extend(args)

    # Start the subprocess and determine the process PID
    subp = subprocess.Popen(final_list)  # starts the process
    pid = subp.pid

    # Wait until the return code returns from the function by considering the timeout. 
    # If not, terminate the process.
    try:
        returncode = subp.wait(timeout=timeout_time)  # should be zero if accomplished
    except subprocess.TimeoutExpired:
        # Distinguish between Linux and Windows and terminate the process if 
        # the timeout has been expired
        if sys.platform == 'linux2':
            os.kill(pid, signal.SIGTERM)
        elif sys.platform == 'win32':
            subp.terminate()

if __name__ == '__main__':
    main()

预先设置Linux命令超时并不是一个糟糕的解决方法,它对我来说是有效的。

cmd = "timeout 20 "+ cmd
subprocess.Popen(cmd.split(), stdout=subprocess.PIPE, stderr=subprocess.PIPE)
(output, err) = p.communicate()

我添加了从jcollado线程到我的Python模块easyprocess的解决方案。

安装:

pip install easyprocess

例子:

from easyprocess import Proc

# shell is not supported!
stdout=Proc('ping localhost').call(timeout=1.5).stdout
print stdout