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! -- https://mail.python.org/mailman/listinfo/python-list
