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print crontab entries for all the users that actually have a crontab

Terminal - print crontab entries for all the users that actually have a crontab
for USER in `cut -d ":" -f1 </etc/passwd`; do crontab -u ${USER} -l 1>/dev/null 2>&1; if [ ! ${?} -ne 0 ]; then echo -en "--- crontab for ${USER} ---\n$(crontab -u ${USER} -l)\n"; fi; done
2009-10-07 20:51:01
User: tharant
Functions: crontab echo
4
print crontab entries for all the users that actually have a crontab

This is how I list the crontab for all the users on a given system that actually have a crontab.

You could wrap it with a function block and place it in your .profile or .bashrc for quick access.

There's prolly a simpler way to do this. Discuss.

Alternatives

There is 1 alternative - vote for the best!

Terminal - Alternatives
for USER in `ls /var/spool/cron`; do echo "=== crontab for $USER ==="; echo $USER; done
2009-10-08 23:51:43
User: jemmille
Functions: echo
-4

Lists crontab for all users on system that have crontabs.

Know a better way?

If you can do better, submit your command here.

What others think

Much shorter:

for USER in /var/spool/cron/*; do echo "--- crontab for $USER ---"; cat $USER; done
Comment by flatcap 299 weeks and 4 days ago

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