On Sat, Jan 27, 2018 at 11:09 AM, Chris Angelico <ros...@gmail.com> wrote: > On Sun, Jan 28, 2018 at 2:58 AM, 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? > > Looks like you're trying to invoke a process without a full path? It > could be because cron jobs execute in a restricted environment. Two > things to try: > > 1) Dump out os.environ to a file somewhere (maybe in /tmp if you don't > have write permission). See if $PATH is set to some really short > value, or at least to something different from what you see when you > run it outside of cron. > > 2) Replace the word "service" with whatever you get from running > "which service" on your system. On mine, it's "/usr/sbin/service". See > what that does. > > Might not solve your problem, but it's worth a try.
Thanks! Using the full path fixed the issue. -- https://mail.python.org/mailman/listinfo/python-list
