W dniu sobota, 27 stycznia 2018 16:59:50 UTC+1 użytkownik larry....@gmail.com napisał: > 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?
Larry, I have exactly the same problem. I'd like to run a script and from normal user it works, and from cron doesn't. In sumarry I was googled to find information how to set two or more env. variables and pass them to subprocess.open. I also try to read $HOME/.profile where usually these env. var. are setting. Have anyone see any example how to do it? Please let me know. Regards, Daniel -- https://mail.python.org/mailman/listinfo/python-list
