How can I programmatically create a new cron job?

Question

I want to be able to programatically add a new cron job, what is the best way to do this?

From my research, it seems I could dump the current crontab and then append a new one, piping that back into crontab:

(crontab -l ; echo "0 * * * * wget -O - -q http://www.example.com/cron.php") | crontab -

Is there a better way?

Answer 1

To Add something to cron

(crontab -l ; echo "0 * * * * hupChannel.sh") 2>&1 | grep -v "no crontab" | sort | uniq | crontab -

To remove this from cron

(crontab -l ; echo "0 * * * * hupChannel.sh") 2>&1 | grep -v "no crontab" | grep -v hupChannel.sh |  sort | uniq | crontab -

hope would help someone

Answer 2

also you can add your tasks to /etc/cron.*/

Answer 3

OP's solution has a bug, it might allow entries to be added twice, use below to fix.

(crontab -l ; echo "0 * * * * your_command") | sort - | uniq - | crontab -

Answer 4

If you're planning on doing it for a run-once scenario for just wget'ing something, take a look at 'at'

Answer 5

Most of the solutions here are for adding lines to the crontab. If you need more control, you'll want to be able to control the entire contents of the crontab.

You can use piping to do this pretty elegantly.

To completely rewrite the crontab, do

echo "2 2 2 2 2 /bin/echo foobar" |crontab -

This should be easy to combine with other answers described here like

crontab -l | <something> | tee |crontab -

Or, if you have the contents in a file, it is even simpler

cat <file> |crontab -

Answer 6

man crontab is also useful:

CRONTAB(1)

NAME

   crontab - manipulate per-user crontabs (Dillon's Cron)

SYNOPSIS

   crontab file [-u user] - replace crontab from file

   crontab - [-u user] - replace crontab from stdin

   crontab -l [user] - list crontab for user