CODEWITHSUNDEEP
pythonlinuxargumentssubprocess
zuberuber
zuberuber

Reputation: 4081

Python subprocess command arguments

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()?

Upvotes: 2

Views: 3639

Answers (2)

amoffat
amoffat

Reputation: 714

You might find sh.py more friendly:

import sh

print sh.ls("-la")

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1121406

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']

Upvotes: 6

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