从 crontab 运行时调用 `arp` 的 Python 脚本会产生错误

Question

我的服务器:

Linux dhcpns 3.19.0-28-generic #30-Ubuntu SMP Mon Aug 31 15:52:51 UTC 2015 x86_64 x86_64 x86_64 GNU/Linux

我的 Python 脚本:

#! /usr/bin/python

import syslog, traceback
import subprocess as sp

def getarp():
  cmd = ["arp", "-a"]
  arp = sp.Popen(cmd, stdout=sp.PIPE, stderr=sp.PIPE)
  output, err = arp.communicate()
  entries = output.splitlines()
  return entries

def syslog_trace(trace):
  log_lines = trace.split('\n')
  for line in log_lines:
    if len(line):
      syslog.syslog(syslog.LOG_ALERT,line)

if __name__ == '__main__':
  try:
    lstArp =  getarp()
    print lstArp
  except Exception as e:
    syslog.syslog(syslog.LOG_ALERT,e.__doc__)
    syslog_trace(traceback.format_exc())
    raise

当我从命令行运行脚本时，它会产生预期的输出。但是，当我从 crontab -e 调用脚本(显然将输出重定向到文件)时，脚本失败。

日志:

Sep 28 18:36:01 dhcpns testarp.py[9984]: OS system call failed.
Sep 28 18:36:01 dhcpns testarp.py[9984]: Traceback (most recent call last):
Sep 28 18:36:01 dhcpns testarp.py[9984]: File "/home/ubuntu/testarp.py", line 23, in <module>
Sep 28 18:36:01 dhcpns testarp.py[9984]: lstArp =  getarp()
Sep 28 18:36:01 dhcpns testarp.py[9984]: File "/home/ubuntu/testarp.py", line 10, in getarp
Sep 28 18:36:01 dhcpns testarp.py[9984]: arp = sp.Popen(cmd, stdout=sp.PIPE, stderr=sp.PIPE)
Sep 28 18:36:01 dhcpns testarp.py[9984]: File "/usr/lib/python2.7/subprocess.py", line 710, in __init__
Sep 28 18:36:01 dhcpns testarp.py[9984]: errread, errwrite)
Sep 28 18:36:01 dhcpns testarp.py[9984]: File "/usr/lib/python2.7/subprocess.py", line 1335, in _execute_child
Sep 28 18:36:01 dhcpns testarp.py[9984]: raise child_exception
Sep 28 18:36:01 dhcpns testarp.py[9984]: OSError: [Errno 2] No such file or directory

什么鬼？我错过了什么？

Answer 1

通过 cron 运行的命令继承不同的环境。特别是，PATH envvar 可能会有所不同。要解决此问题，请在 Python 脚本中指定 arp 程序的完整路径，或在相应的 crontab 中显式配置 PATH:

PATH=/bin:/usr/bin:/path/where/arp/lives