error from Popen only when run from cron
Wildman
best_lay at yahoo.com
Sat Jan 27 11:52:23 EST 2018
On Sat, 27 Jan 2018 10:58:36 -0500, Larry Martell wrote:
> I have a script that does this:
>
> subprocess.Popen(['service', 'some_service', 'status'],
> stdout=subprocess.PIPE, stderr=subprocess.STDOUT)
>
> When I run it from the command line it works fine. When I run it from
> cron I get:
>
> subprocess.Popen(['service', 'some_service', 'status'],
> stdout=subprocess.PIPE, stderr=subprocess.STDOUT)
> 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
>
> Anyone have any clue as to what file it's complaining about? Or how I
> can debug this further?
Cron provides this as $PATH: /usr/bin;/usr/sbin
>From within a terminal enter: whereis service
If service is not in Cron's $PATH, that is your problem.
Adding the complete path to 'service' in the script
should fix things.
If service is in Cron's $PATH, I have no further ideas.
--
<Wildman> GNU/Linux user #557453
The cow died so I don't need your bull!
More information about the Python-list
mailing list