How do you do a simple "chmod +x" from within python?

141

votes

I want to create a file from within a python script that is executable.

import os
import stat
os.chmod('somefile', stat.S_IEXEC)

it appears os.chmod doesn't 'add' permissions the way unix chmod does. With the last line commented out, the file has the filemode -rw-r--r--, with it not commented out, the file mode is ---x------. How can I just add the u+x flag while keeping the rest of the modes intact?

pythonchmod

229

votes

Use os.stat() to get the current permissions, use | to or the bits together, and use os.chmod() to set the updated permissions.

Example:

import os
import stat

st = os.stat('somefile')
os.chmod('somefile', st.st_mode | stat.S_IEXEC)

24

votes

For tools that generate executable files (e.g. scripts), the following code might be helpful:

def make_executable(path):
    mode = os.stat(path).st_mode
    mode |= (mode & 0o444) >> 2    # copy R bits to X
    os.chmod(path, mode)

This makes it (more or less) respect the umask that was in effect when the file was created: Executable is only set for those that can read.

Usage:

path = 'foo.sh'
with open(path, 'w') as f:           # umask in effect when file is created
    f.write('#!/bin/sh\n')
    f.write('echo "hello world"\n')

make_executable(path)

16

votes

If you know the permissions you want then the following example may be the way to keep it simple.

Python 2:

os.chmod("/somedir/somefile", 0775)

Python 3:

os.chmod("/somedir/somefile", 0o775)

Compatible with either (octal conversion):

os.chmod("/somedir/somefile", 509)

reference permissions examples

8

votes

If you're using Python 3.4+, you can use the standard library's convenient pathlib.

Its Path class has built-in chmod and stat methods.

from pathlib import Path
import stat


f = Path("/path/to/file.txt")
f.chmod(f.stat().st_mode | stat.S_IEXEC)

4

votes

You can also do this

>>> import os
>>> st = os.stat("hello.txt")

Current listing of file

$ ls -l hello.txt
-rw-r--r--  1 morrison  staff  17 Jan 13  2014 hello.txt

Now do this.

>>> os.chmod("hello.txt", st.st_mode | 0o111)

and you will see this in the terminal.

ls -l hello.txt    
-rwxr-xr-x  1 morrison  staff  17 Jan 13  2014 hello.txt

You can bitwise or with 0o111 to make all executable, 0o222 to make all writable, and 0o444 to make all readable.

4

votes

Respect umask like chmod +x

man chmod says that if augo is not given as in:

chmod +x mypath

then a is used but with umask:

A combination of the letters ugoa controls which users' access to the file will be changed: the user who owns it (u), other users in the file's group (g), other users not in the file's group (o), or all users (a). If none of these are given, the effect is as if (a) were given, but bits that are set in the umask are not affected.

Here is a version that simulates that behavior exactly:

#!/usr/bin/env python3

import os
import stat

def get_umask():
    umask = os.umask(0)
    os.umask(umask)
    return umask

def chmod_plus_x(path):
    os.chmod(
        path,
        os.stat(path).st_mode |
        (
            (
                stat.S_IXUSR |
                stat.S_IXGRP |
                stat.S_IXOTH
            )
            & ~get_umask()
        )
    )

chmod_plus_x('.gitignore')

Tested in Ubuntu 16.04, Python 3.5.2.

-1

votes

In python3:

import os
os.chmod("somefile", 0o664)

Remember to add the 0o prefix since permissions are set as an octal integer, and Python automatically treats any integer with a leading zero as octal. Otherwise, you are passing os.chmod("somefile", 1230) indeed, which is octal of 664.

How do you do a simple "chmod +x" from within python?

7 Answers