Cannot find the file specified when using subprocess.call('dir', shell=True) in Python

path, python, python-2.7, shell, subprocess

Solution

I think you may have a problem with your `COMSPEC` environment variable:

>>> import os
>>> os.environ['COMSPEC']
'C:\\Windows\\system32\\cmd.exe'
>>> import subprocess
>>> subprocess.call('dir', shell=True)

    (normal output here)

>>> os.environ['COMSPEC'] = 'C:\\nonexistent.exe'
>>> subprocess.call('dir', shell=True)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "c:\Python27\lib\subprocess.py", line 493, in call
    return Popen(*popenargs, **kwargs).wait()
  File "c:\Python27\lib\subprocess.py", line 679, in __init__
    errread, errwrite)
  File "c:\Python27\lib\subprocess.py", line 896, in _execute_child
    startupinfo)
WindowsError: [Error 2] The system cannot find the file specified

I discovered this potential issue by digging into `subprocess.py` and looking in the `_execute_child` function, as pointed-to by the traceback. There, you'll find a block starting with `if shell:` that will search the environment for said variable and use it to create the arguments used to launch the process.

Problem

In a 64-bit system with 32 bit python 2.7 installed I am trying to do the following: ``` import subprocess p = subprocess.call('dir', shell=True) print p ``` But this gives me: ``` Traceback (most recent call last): File "test.py", line 2, in <module> p = subprocess.call('dir', shell=True) File "C:\Python27\lib\subprocess.py", line 522, in call return Popen(*popenargs, **kwargs).wait() File "C:\Python27\lib\subprocess.py", line 709, in __init__ errread, errwrite) File "C:\Python27\lib\subprocess.py", line 957, in _execute_child startupinfo) WindowsError: [Error 2] The system cannot find the file specified ``` If I in the terminal do... ``` dir ``` ...it of course prints the present folder content. I have tried to change the shell parameter to shell=False. Edit: Actually I cannot call any executable on the path with `subprocess.call()`. The statement `p = subprocess.call('dir', shell=True)` works fine on another machine and I think that it is related. If I do ``` subprocess.call('PATH', shell=True) ``` then I get ``` Traceback (most recent call last): File "test.py", line 4, in <module> subprocess.call('PATH', shell=True) File "C:\Python27\lib\subprocess.py", line 522, in call return Popen(*popenargs, **kwargs).wait() File "C:\Python27\lib\subprocess.py", line 709, in __init__ errread, errwrite) File "C:\Python27\lib\subprocess.py", line 957, in _execute_child startupinfo) WindowsError: [Error 2] The system cannot find the file specified ``` If I do: ``` import os print os.curdir ``` then I get ``` . ``` All of the above is executed in the terminal started in Administrator mode.

Original source