Python - 创建超时(终止子进程)函数的问题

0 投票
3 回答
619 浏览
提问于 2025-04-17 09:40

我需要你们的帮助,我在写一个小程序,这个程序会创建一个子进程,执行10次ping命令,并且应该在给定的超时时间(5秒)后被终止。但这个例子没有成功。你能给我点建议吗?

谢谢。

斯特凡

输出:

Traceback (most recent call last):
File "./nwcheck.py.work", line 146, in <module>
MyCheck().check().exit()
File "./nwcheck.py.work", line 80, in check
output = process.communicate()
File "/usr/lib/python2.6/subprocess.py", line 701, in communicate
return self._communicate(input)
File "/usr/lib/python2.6/subprocess.py", line 1199, in _communicate
rlist, wlist, xlist = select.select(read_set, write_set, [])
File "./nwcheck.py.work", line 29, in alarm_handler
raise alarm
TypeError: exceptions must be old-style classes or derived from BaseException, not   builtin_function_or_method

代码:

def check(self):

    class Alarm(Exception):
        pass

    def alarm_handler(signum, frame):
        raise alarm

    def get_process_children(pid):
        p = Popen('ps --no-headers -o pid --ppid %d' % pid, shell = True,
                   stdout = PIPE, stderr = PIPE)
        stdout, stderr = p.communicate()
        return [int(p) for p in stdout.split()]
    timeout = 5
    args2 = [
            'ping',
            'localhost',
            '-c 10',
            ]
    process = subprocess.Popen(args2, stdout=subprocess.PIPE, stderr=subprocess.PIPE, env={'LANG':'de_DE@euro'})
    processpid = process.pid
    print processpid
    if timeout != -1:
        signal(SIGALRM, alarm_handler)
        alarm(timeout)
        print 'in timeout abfrage'
    try:
        ## catch stdout and stderr
        output = process.communicate()
        if timeout != -1:
            alarm(0)
            print 'in timeout abfrage 2'
    except Alarm:
        pids = [process.pid]
        print pids
        if kill_tree:
            pids.extend(get_process_children(process.pid))
        for pid in pids:
        # process might have died before getting to this line
        # so wrap to avoid OSError: no such process
            try:
                kill(pid, SIGKILL)
            except OSError:
                pass
        return -9, '', ''

    # Return a response
    return output

3 个回答

0

一个子进程可以通过以下方式被终止:

import os
os.kill(process.pid, signal.SIGKILL)

或者可能是:

from subprocess import Popen
Popen.kill()

或者:

from subprocess import Popen
Popen.terminate()

可以查看这个链接了解更多信息: http://docs.python.org/library/subprocess.html#popen-objects

0

你的异常叫做 Alarm,但是你抛出的却是 alarm。在Python中,大小写是有区别的。

你可能想把 Alarm 改个更好理解的名字。像 AlarmException 或者 AlarmError 这样的名字会让代码更清晰。

2

你可能应该用异常处理,而不是用警报:

class AlarmException(Exception): 
    pass 

...

    def alarm_handler(signum, frame):
        raise AlarmException()

另外,别忘了管道可能会出现溢出。如果进程产生的数据太多(在某些Linux系统上超过64k),那么程序就会被卡住。

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