如何使用 python 在 *nix 上获取文件的权限掩码,如 644 或 755?
有没有这样做的功能或类?非常感谢!
如何使用 python 在 *nix 上获取文件的权限掩码,如 644 或 755?
有没有这样做的功能或类?非常感谢!
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'
从这里您将识别典型的八进制权限。
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
你真的只对低位感兴趣,所以你可以砍掉其余的:
>>> 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)
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)
os.lstat
如果文件是符号链接,该函数将为您提供链接本身的模式,而os.stat
取消引用链接。因此,我发现os.lstat
最普遍有用的。
这是一个示例,给定常规文件“testfile”和指向后者的符号链接“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))
该脚本为我输出以下内容:
0777
0666
如果您不想弄清楚 stat 的含义,另一种方法是使用 os.access 命令http://docs.python.org/library/os.html#os.access 但是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
oct(os.stat('file').st_mode)[4:]
os.access(path, mode)
True
如果路径上允许访问,则方法返回,否则返回False
。
可用模式有:
例如,检查文件 /tmp/test.sh 是否具有执行权限
ls -l /tmp/temp.sh
-rw-r--r-- 1 * * 0 Mar 2 12:05 /tmp/temp.sh
os.access('/tmp/temp.sh',os.X_OK)
False
after changing the file permission to +x
chmod +x /tmp/temp.sh
ls -l /tmp/temp.sh
-rwxr-xr-x 1 * * 0 Mar 2 12:05 /tmp/temp.sh
os.access('/tmp/temp.sh',os.X_OK)
True
我确定 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。
这是检查目录权限的简单方法。
import os
import stat
mode = os.stat("path_of_directory").st_mode
if not ((mode & stat.S_IWUSR):
print('not writable by user')
if not ((mode & stat.S_IWUSR) and (mode & stat.S_IWGRP) and (mode & stat.S_IWOTH)):
print('not writable by all')
标志列表如下:
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
os.stat
类似于 c-lib stat (man 2 stat on linux 查看信息)
stats = os.stat('file.txt')
print(stats.st_mode)
如果需要,您可以使用 Popen 运行 Bash stat 命令:
普通的 Bash 命令:
jlc@server:~/NetBeansProjects/LineReverse$ stat -c '%A %a %n' revline.c
-rw-rw-r-- 664 revline.c
然后使用 Python:
>>> from subprocess import Popen, PIPE
>>> fname = 'revline.c'
>>> cmd = "stat -c '%A %a %n' " + fname
>>> out = Popen(cmd, shell=True, stdout=PIPE).communicate()[0].split()[1].decode()
>>> out
'664'
如果您想搜索目录,还有另一种方法:
>>> from os import popen
>>> cmd = "stat -c '%A %a %n' *"
>>> fname = 'revline.c'
>>> for i in popen(cmd):
... p, m, n = i.split()
... if n != fname:
... continue
... print(m)
break
...
664
>>>