with a shell script, it is possible to tell when it's run with cron vs run manually?

with a shell script, it is possible to tell when it's run with cron vs run manually?

UPDATE:

it was asked why I want to know. My cron job will be logging, and i want to be able to log if someone is executing it manually.

Solution 1:

In the times that I've needed to do this, i know it's a script that I'll never redirect to a file or to a pipe. So a simple test is to check if stdout (a.k.a file descriptor 1) is a tty (which it won't be from cron). In bash:

if [ -t 1 ]
then
    : # running from terminal
else
    : # not running from terminal, cron maybe
fi

Again, warning, this is a test simply if your stdout is a tty. But works for my simple purposes.

You could also check what your parent process is. In Linux, it can be as simple as:

if grep -q cron /proc/$PPID/cmdline &> /dev/null
then
    : # running from cron
fi

Solution 2:

You could define a unique environment variable in your crontab:

RUNNINGFROMCRON=1

16 18 * * *     /usr/local/bin/crontest

I tested with this script:

#! /bin/bash

echo " running script "

echo -n "testing for var: "
echo $RUNNINGFROMCRON

When I ran from cron output was:

running script 
testing for var: 1

When I ran from the command line manualy, output was:

running script 
testing for var:

Solution 3:

You can inspect the name of the parent process, which can be retrieved with ps -p $PPID -o comm=. This is spoofable, of course.

However this is not necessarily a good idea (for example, you won't be able to easily test your cron jobs manually). Maybe if you explain why you want this someone can suggest a better solution.