如何获取文件的权限掩码？

Question

如何使用python在*nix上获取文件的权限掩码,如644或755？这样做有什么功能或类吗？你能帮助我吗？非常感谢你!

Answer 1

os.stat是stat(2)系统调用接口的包装器.

>>> import os
>>> from stat import *
>>> os.stat("test.txt") # returns 10-tupel, you really want the 0th element ...
posix.stat_result(st_mode=33188, st_ino=57197013, \
    st_dev=234881026L, st_nlink=1, st_uid=501, st_gid=20, st_size=0, \
    st_atime=1300354697, st_mtime=1300354697, st_ctime=1300354697)

>>> os.stat("test.txt")[ST_MODE] # this is an int, but we like octal ...
33188

>>> oct(os.stat("test.txt")[ST_MODE])
'0100644'

从这里你将认识到典型的八进制权限.

S_IRWXU 00700   mask for file owner permissions
S_IRUSR 00400   owner has read permission
S_IWUSR 00200   owner has write permission
S_IXUSR 00100   owner has execute permission
S_IRWXG 00070   mask for group permissions
S_IRGRP 00040   group has read permission
S_IWGRP 00020   group has write permission
S_IXGRP 00010   group has execute permission
S_IRWXO 00007   mask for permissions for others (not in group)
S_IROTH 00004   others have read permission
S_IWOTH 00002   others have write permission
S_IXOTH 00001   others have execute permission

你真的只有在较低的兴趣位,所以你可以砍掉的其余部分:

>>> oct(os.stat("test.txt")[ST_MODE])[-3:]
'644'
>>> # or better
>>> oct(os.stat("test.txt").st_mode & 0o777)

旁注:上部确定文件类型,例如:

S_IFMT  0170000 bitmask for the file type bitfields
S_IFSOCK    0140000 socket
S_IFLNK 0120000 symbolic link
S_IFREG 0100000 regular file
S_IFBLK 0060000 block device
S_IFDIR 0040000 directory
S_IFCHR 0020000 character device
S_IFIFO 0010000 FIFO
S_ISUID 0004000 set UID bit
S_ISGID 0002000 set-group-ID bit (see below)
S_ISVTX 0001000 sticky bit (see below)

Answer 2

我认为这是获取文件权限位的最明确方式:

stat.S_IMODE(os.lstat("file").st_mode)

os.lstat函数,如果文件是符号链接,将为您提供链接本身的模式,而os.stat取消引用该链接.因此我发现os.lstat最常用.

这是一个示例案例,给定常规文件"testfile"和符号链接到后者,"testlink":

import stat
import os

print oct(stat.S_IMODE(os.lstat("testlink").st_mode))
print oct(stat.S_IMODE(os.stat("testlink").st_mode))

这个脚本为我输出以下内容:

0777
0666

Answer 3

另一种方法是,如果你不想弄清楚stat的意思是使用os.access命令http://docs.python.org/library/os.html#os.access 但是请阅读有关可能的文档安全问题

例如,检查具有读/写权限的文件test.dat的权限

os.access("test.dat",os.R_OK)
>>> True

#Execute permissions
os.access("test.dat",os.X_OK)
>>> False

#And Combinations thereof
os.access("test.dat",os.R_OK or os.X_OK)
>>> True

os.access("test.dat",os.R_OK and os.X_OK)
>>> False

Answer 4

我确信 os 模块中有很多基于文件的函数。如果你运行，os.stat(filename)你总是可以解释结果。

http://docs.python.org/library/stat.html