error from Popen only when run from cron

[Larry Martell]

On Sat, Jan 27, 2018 at 11:09 AM, Chris Angelico <rosuav at gmail.com> wrote:
> On Sun, Jan 28, 2018 at 2:58 AM, Larry Martell <larry.martell at 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.