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?

Answer 1

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.

Answer 2

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.