Python subprocess command arguments

Question

Why if I run subprocess.check_output('ls') everything is working but when I add argument to command like: subprocess.check_output('ls -la') I get error:

Traceback (most recent call last):
  File "", line 1, in 
  File "/usr/lib/python2.7/subprocess.py", line 537, in check_output
    process = Popen(stdout=PIPE, *popenargs, **kwargs)
  File "/usr/lib/python2.7/subprocess.py", line 679, in __init__
    errread, errwrite)
  File "/usr/lib/python2.7/subprocess.py", line 1259, in _execute_child
    raise child_exception
OSError: [Errno 2] No such file or directory

How can I pass command arguments into subprocess.check_output()?

Martijn Pieters · Accepted Answer · 2012-11-11 14:30:21Z

6

You need to split the arguments into a list:

subprocess.check_output(['ls', '-la'])

The subprocess callables do not parse the command out to individual arguments like the shell does. You either need to do this yourself or you need to tell subprocess to use the shell explicitly:

subprocess.check_output('ls -la', shell=True)

The latter is not recommended as it can expose your application to security vulnerabilities. You can use shlex.split() to parse a shell-like command line if needed:

>>> import shlex
>>> shlex.split('ls -la')
['ls', '-la']

answered Nov 11, 2012 at 14:30

Martijn Pieters

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Comments

amoffat · Accepted Answer · 2012-11-12 05:56:21Z

0

You might find sh.py more friendly:

import sh

print sh.ls("-la")

answered Nov 12, 2012 at 5:56

amoffat

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