[FIXED] Unable to capture result of ls -la with subprocess.Popen

Issue

I am trying to capture the output when I execute a custom command using Popen:

import subprocess

def exec_command():
    command = "ls -la" # will be replaced by my custom command 
    result = subprocess.Popen(command, stdout=subprocess.PIPE).communicate()[0]
    print(result)

exec_command()

I get an OSError with following stacktrace:

 File "/usr/lib64/python2.7/subprocess.py", line 711, in __init__
    errread, errwrite)
  File "/usr/lib64/python2.7/subprocess.py", line 1327, in _execute_child
    raise child_exception
OSError: [Errno 2] No such file or directory

Please let me know what I would need to use.

Note: The stacktrace shows the code was executed in Python 2.7, but I got the same error running with Python 2.6

Solution

When running without shell=True (which you are doing, correctly; shell=True is dangerous), you should pass your command as a sequence of the command and the arguments, not a single string. Fixed code is:

def exec_command():
    command = ["ls", "-la"]  # list of command and argument
    ... rest of code unchanged ...

If you had user input involved for some of the arguments, you'd just insert it into the list:

def exec_command(somedirfromuser):
    command = ["ls", "-la", somedirfromuser]

Note: If your commands are sufficiently simple, I'd recommend avoiding subprocess entirely. os.listdir and os.stat (or on Python 3.5+, os.scandir alone) can get you the same info as ls -la in a more programmatically usable form without the need to parse it, and likely faster than launching an external process and communicating with it via a pipe.

Answered By - ShadowRanger