Question

我想要一个给出目录的python程序，它将返回该目录中具有775（rwxrwxr-x）权限的所有目录

谢谢！

Answer 1

这两个答案都没有得到解决，尽管并不完全清楚这是OP想要的。这是一种递归方法（未经测试，但你明白了）：

import os
import stat
import sys

MODE = "775"

def mode_matches(mode, file):
    """Return True if 'file' matches 'mode'.

    'mode' should be an integer representing an octal mode (eg
    int("755", 8) -> 493).
    """
    # Extract the permissions bits from the file's (or
    # directory's) stat info.
    filemode = stat.S_IMODE(os.stat(file).st_mode)

    return filemode == mode

try:
    top = sys.argv[1]
except IndexError:
    top = '.'

try:
    mode = int(sys.argv[2], 8)
except IndexError:
    mode = MODE

# Convert mode to octal.
mode = int(mode, 8)

for dirpath, dirnames, filenames in os.walk(top):
    dirs = [os.path.join(dirpath, x) for x in dirnames]
    for dirname in dirs:
        if mode_matches(mode, dirname):
            print dirname

stdlib文档中描述了类似的东西 stat

Answer 2

查看os模块。特别是os.stat来查看权限位。

import  os

for filename in os.listdir(dirname):
   path=os.path.join(dirname, filename)
   if os.path.isdir(path):
       if (os.stat(path).st_mode & 0777) == 0775:
           print path

Answer 3

紧凑型发电机基于Brian的回答：

import os

(fpath for fpath 
   in (os.path.join(dirname,fname) for fname in os.listdir(dirname)) 
   if (os.path.isdir(fpath) and (os.stat(fpath).st_mode & 0777) == 0775))

Answer 4

它必须是python吗？

你也可以使用find来做到这一点：

“找。-perm 775”

Answer 5

您可以使用以下命令检查775文件和目录的权限

ValueError: Wrong number of columns at line 2

检查python中目录的权限

5 个答案: