Pythonic 方式来遍历阴影等

Question

我正在尝试使用 python 实现逻辑:

 cat /etc/shadow | awk -F: '($2 == "" ) { print $1 " does not have a password "}'

如果上面返回用户的输出，我会这样做

passwd -l <username>

我正在尝试使用 python 实现上述逻辑，但我不太确定它是否能以这种方式工作；这是我的Python代码:

/etc/shadow 看起来像

root:*:17709:0:99999:7:::
daemon:*:17709:0:99999:7:::
bin:*:17709:0:99999:7:::
sys:*:17709:0:99999:7:::
sync:*:17709:0:99999:7:::
games:*:17709:0:99999:7:::
man:*:17709:0:99999:7:::
lp:*:17709:0:99999:7:::
mail:*:17709:0:99999:7:::
news:*:17709:0:99999:7:::
uucp:*:17709:0:99999:7:::
proxy:*:17709:0:99999:7:::
www-data:*:17709:0:99999:7:::
backup:*:17709:0:99999:7:::

代码

with open("/etc/shadow") as file:
   for line in file:
       line = line.rstrip()
       if line[line.find(":")+1:line.find(":")]=="":
          print "This is a problem"
       elif line[line.find(":")+1:line.find(":")]=="*":
          print line[line.find(":")+1:line.find(":")]
       else:
          print "All Good"

上面的代码返回“这是一个问题”，这是不对的

Answer 1

您可以使用re提取所需的列:

import re

data = """root:*:17709:0:99999:7:::
daemon:*:17709:0:99999:7:::
bin:*:17709:0:99999:7:::
sys:*:17709:0:99999:7:::
sync:*:17709:0:99999:7:::
games:*:17709:0:99999:7:::
man:*:17709:0:99999:7:::
lp:*:17709:0:99999:7:::
mail:*:17709:0:99999:7:::
news:*:17709:0:99999:7:::
uucp:*:17709:0:99999:7:::
proxy:*:17709:0:99999:7:::
www-data:*:17709:0:99999:7:::
backup:*:17709:0:99999:7:::"""

groups = re.findall('(.*?):(.*?):(.*?):(.*?):(.*?):(.*?):(.*?):(.*?):', data)

if all(g[1].strip() for g in groups):
    print('All good')
else:
    print('This is a problem')

打印:

All good

此正则表达式的说明 here 。在第二组 (g[1]) 中，您拥有隐藏密码 (*) 或空字符串。