如何获取文件的权限掩码？

How can I get a file's permission mask like 644 or 755 on *nix using python? Is there any function or class for doing that? Could you guys help me out? Thank you very much!

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！

6 个解决方案

#1

os.stat is a wrapper around the stat(2) system call interface.

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'

From here you'll recognize the typical octal permissions.

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

You are really only interested in the lower bits, so you could chop off the rest:

你真的只对低位感兴趣，所以你可以砍掉其余部分：

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

Sidenote: the upper parts determine the filetype, e.g.:

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)

#2

I think this is the clearest way of getting a file's the permission bits:

我认为这是获取文件权限位的最清晰方法：

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

The os.lstat function, will in case the file is a symlink, give you the mode of the link itself, whereas os.stat dereferences the link. Therefore I find os.lstat the most generally useful.

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。

Here's an example case, given regular file "testfile" and symlink to the latter, "testlink":

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))

This script outputs the following for me:

这个脚本为我输出以下内容：

0777
0666

#3

Another way to do it if you don't want to work out what stat means is to use the os.access command http://docs.python.org/library/os.html#os.access BUT read the docs about possible security issues

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

For instance to check permissions on the file test.dat which has read/write permissions

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

#4

oct(os.stat('file').st_mode)[4:]

辛（os.stat（ '文件'）ST_MODE。）[4：]

#5

There are a lot of file based functions inside the os module im sure. If you run os.stat(filename) you can always interprate the results.

os模块中有很多基于文件的功能。如果运行os.stat（filename），则始终可以插入结果。

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

#6

os.stat is analogous to the c-lib stat (man 2 stat on linux to see the information)

os.stat类似于c-lib stat（linux上的man 2 stat查看信息）

stats = os.stat('file.txt')
print stats.st_mode

#1