One liner that randomly rewrites the startup time of cron

It is a small story.

Prerequisite knowledge

In bash, you can randomly get numbers from 0 to 32767 by using the environment variable $ RANDOM.

For example, to randomly display numbers from 0 to 3, combine the modulo operator symbols % as follows:

Randomly display numbers from 0 to 3

echo $((RANDOM % 4))

This is the number from 5 to 59.

Randomly display numbers from 5 to 59

echo $((5 + RANDOM % 55))

The explanation of this area was much easier to understand in [Randomly do it with a shell script], so please refer to that.

To rewrite the cron time field

The main subject. Suppose the following cron is defined on the last line:

00 2 * * * /home/mindwood/foo.sh > /dev/null 2>&1

It is set to start at 2:00 am, but I would like to set this to start from 0:05 am to 3:59 am and give it a range. On top of that, I want to avoid just every hour (1 am, 2 am ...).

Here is the one liner to rewrite.

H=$((RANDOM%4));M=$((5+RANDOM%55));sed -i -e '$s/^[0-9]\+ [0-9]\+/'$M' '$H'/' /var/spool/cron/crontabs/root

For example, it is updated as follows.

43 1 * * * /home/mindwood/foo.sh > /dev/null 2>&1

If you don't like this time, you can start over again.

Note that $ in sed represents the last line. For more information on sed, please refer to [How to write in this case with sed?].