Error running a program using subprocess

Question

I have a tool which returns me some info about the machine I am running on.On the normal command line it would be something like -

sudo /path-to-tool-directory/tool arg

and this works fine .Now when I break this up and include this in my python script as

result = subprocess.call (["sudo /path-to-tool-directory/tool","arg"])

it throws me an error

subprocess.py in line XYZ ,
in _execute_child
raise child_exception
OSError: [Errno 2] No such file or directory

any clue what might be going wrong here?

zzzirk · Accepted Answer · 2012-10-12 05:58:34Z

6

When using the subprocess module you need to provide the call() function with a list of command line arguments. Taking your example above:

result = subprocess.call (["sudo /path-to-tool-directory/tool","arg"])

This won't work because "sudo /path-to-tool-directory/tool" is a single list item. What you need is all items to be individual list items:

result = subprocess.call (["sudo", "/path-to-tool-directory/tool", "arg"])

This should successfully run and terminate leaving the return code from sudo in result.

answered Oct 12, 2012 at 5:58

zzzirk

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1 Comment

hughdbrown Over a year ago

+1 Nice answer. Gives the reasons for OP's problem pretty clearly.

hughdbrown · Accepted Answer · 2012-10-12 14:33:28Z

0

Split off the call to sudo (for all the reasons that @zzzrik elaborates on above):

>>> result = subprocess.call (["sudo /usr/bin/python","/home/hughdbrown/Dropbox/src/longwords.py"])
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/subprocess.py", line 493, in call
    return Popen(*popenargs, **kwargs).wait()
  File "/usr/lib/python2.7/subprocess.py", line 679, in __init__
    errread, errwrite)
  File "/usr/lib/python2.7/subprocess.py", line 1249, in _execute_child
    raise child_exception
OSError: [Errno 2] No such file or directory
>>> result = subprocess.call (["sudo", "/usr/bin/python","/home/hughdbrown/Dropbox/src/longwords.py"])
[sudo] password for hughdbrown:

See? The second one is working because I get prompted for a password.

edited Oct 12, 2012 at 14:33

answered Oct 12, 2012 at 1:33

hughdbrown

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3 Comments

Manish Over a year ago

nope doesnt work for me .Infact I made it even simpler and I tried just this in my file.py "g = subprocess.call([ "/usr/lib/vim"])" and that gave me the same exception

hughdbrown Over a year ago

Is vim actually in /usr/lib? Mine is in /usr/bin.

hughdbrown Over a year ago

I would start by calling which vim (or whatever you want to test) and then putting that exact path as the first argument to subprocess.call. That way, you know that you will not get "no such file or directory".

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