Is subprocess.Popen completely broken?

[Gary Herron]

Skip Montanaro wrote:
>>>  >>> proc = subprocess.Popen ("ls /tmp")
>>>       
>> proc = subprocess.Popen ("ls /tmp", shell=True)
>>
>> or
>>
>> proc = subprocess.Popen (["ls", "/tmp"])
>>
>> should work.
>>     
>
> Why should I need to set shell=True?  I'm not globbing anything.  The
> second case still fails:
>
>   
>>>> proc = subprocess.Popen (["/usr/bin/ls" "/tmp"])
>>>>         
This line fails because of a typo.  There needs to be a comma between 
the two list elements.  Without the comma, this becomes

    proc = subprocess.Popen (["/usr/bin/ls/tmp"])

which obviously fails.

Gary Herron


> Traceback (most recent call last):
>   File "<stdin>", line 1, in ?
>   File "/opt/app/g++lib6/python-2.4/lib/python2.4/subprocess.py", line 542, in
> __init__
>     errread, errwrite)
>   File "/opt/app/g++lib6/python-2.4/lib/python2.4/subprocess.py", line 975, in
> _execute_child
>     raise child_exception
> OSError: [Errno 20] Not a directory
>
> Thx,
>
> Skip
>
>
>