Question

我想找出'ps'命令的输出是否包含进程'smtpd' 问题是各种busybox需要不同的ps命令！有些需要'ps x'，其他需要'ps w'而其他只需'ps'

我如何制作一个尝试所有'ps'可能性的通用算法？

示例：

linex=''
foo=os.popen('ps')
for x in foo.readlines():
   if x.lower().find('smtpd') != -1:
     // SOME SCRIPT STUFF on linex string...
return linex

linex=''
foo=os.popen('ps w')
for x in foo.readlines():
   if x.lower().find('smtpd') != -1:
     // SOME SCRIPT STUFF on linex string...
return linex

linex=''
foo=os.popen('ps x')
for x in foo.readlines():
   if x.lower().find('smtpd') != -1:
     // SOME SCRIPT STUFF on linex string...
return linex

Answer 1

检查一下：

Process list on Linux via Python

/ proc是您找到所需内容的正确位置

import os

pids = [pid for pid in os.listdir('/proc') if pid.isdigit()]

for pid in pids:
    try:
        cmd = open(os.path.join('/proc', pid, 'cmdline'), 'rb').read()
        if cmd.find('smtpd') != -1:
            print "PID: %s; Command: %s" % (pid, cmd)
    # process has already terminated
    except IOError:
        continue

Answer 2

def find_sys_cmds(needle,cmd,options):
    for opt in options:
        for line in os.popen("%s %s"%(cmd,opt)).readlines():
            if needle in line.lower():
                return line

print find_sys_cmds("smtpd","ps",["","x","w","aux",..."])

是你可以这样做的一种方式

如果您可能有多个匹配流程

def find_sys_cmds(needle,cmd,options):
     for opt in options:
         for line in os.popen("%s %s"%(cmd,opt)).readlines():
             if needle in line.lower():
                    yield line

for line in find_sys_cmds("smtpd","ps",["","x","w","aux",..."]):
    print line

如何在Python中找到unix进程

2 个答案: