Getting subprocess.Popen stdout when running by cron

Question

I want to get a service status and if it's not up, to send the status (stdout) in email.
This script is scheduled to run every hour by cron.
When running manually, the following works fine:

def is_service_running(name):
with open(os.devnull, 'wb') as hide_output:
    proc = subprocess.Popen(['service', name, 'status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)
        output = proc.stdout.read()
        exit_code = proc.wait()
        return exit_code == 0, output

But when running by cron. output is empty.
How can I capture stdout when running by cron? Thank you

Is it because /usr/sbin is in your interactive environment but not cron’s? — Jeff Schaller
– Jeff Schaller ♦, Commented Jun 14, 2018 at 17:58
I think you should see if that popen call is successful. Maybe also use the full path to the service command. — Jeff Schaller
– Jeff Schaller ♦, Commented Jun 15, 2018 at 9:21

SagiLow · Accepted Answer · 2018-06-15 09:22:15Z

1

The problem wasn't cron but shell=True.
Apparently, when using shell=True, popen expects single string and not a list.
So when I updated my call to:

proc = subprocess.Popen(['service ' + name + ' status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)

everything worked.

answered Jun 15, 2018 at 9:22

SagiLow

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