On 2018-01-27, Larry Martell <larry.mart...@gmail.com> 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?
Try using the complete path of the executable. Cron jobs run with a pretty limited set of environment variables and may not have the PATH value you expect. -- Grant -- https://mail.python.org/mailman/listinfo/python-list
