How can I get the Unix permission mask from a file? [duplicate]

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? Thank you very much!

Solution 1:

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

>>> 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.

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:

>>> 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.:

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)

Solution 2:

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

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.

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

0777
0666

Solution 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

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

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

Solution 4:

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