Cron

[Jose Mosquera]

I need to differ when a script is executed by the cron. How can I identify inside a script that the cron is who executes it?

Rgds.

[Pete Randall]

Have cron invoke a wrapper script which sets the FROMCRON variable and then calls the common script.


Pete

[Peter Godron]

Hi,
variation on Pete's suggestion:

Modify the actual cron call:

05 12 * * * export FROMCRON="cron";/a.sh > /a.lis 2>&1

[john korterman · Accepted solution]

Hi Jose,

I tried to insert this at the beginning of a script:

#!/usr/bin/sh

PARENT=$(UNIX95= ps -ef | grep $PPID | awk -v ppid=$PPID '$2 ~ ppid {print $3}')

CRON=$(UNIX95= ps -C cron | grep -v CMD| awk '{print $1}')

if [ $PARENT = $CRON ]
then
echo executed by cron
else
echo not executed by cron
fi

It seems to work on my system, but may not work anywhere else!


regards,
John K.

[A. Clay Stephenson]

Two ways immediately suggest themselves:

1) Link (hard) the normal version of the script and a cron version of the script.

e.g. normalscript.sh crnvrsionXY123.sh

Now your code simply tests ${0} to see which name was used to invoke the script.
Note that you intentionally make the cron'ed name difficult to use by including
deliberately obtuse characters.

2) Test stdin to see if it is a TTY device (ie, a terminal)

if [ -t 0 ]
then
echo "Interactive"
else
echo "Non-interactive; probably cron"
fi

[spex]

Jose,

I found this on the forums some time ago:

typeset INTERACTIVE=0 && [ -t 0 ] && INTERACTIVE=1

Put it in the beginning of the script to be executed, and then test 'INTERACTIVE'.

PCS

[Jose Mosquera]

Thank you very much!