[SOLVED] Add To Crontab (if not already exists) Using Bash Script

Issue

What is the correct way for a bash script to add a job to crontab, such that

• there will be no duplicate jobs
• the crontab file will not be resorted
• (optional) close to being a one-liner

Came across this solution below, but it does not affect the output of running crontab -l.

grep 'some_user python /mount/share/script.py' /etc/crontab || echo '*/1 *  *  *  * some_user python /mount/share/script.py' >> /etc/crontab

Tried converting it to affect crontab -l,

(crontab -l | grep '/mount/share/script.py') || { crontab -l; '*/1 * * * * some_user python /mount/share/script.py >> /root/foo/logs/foo.cron.log 2>&1'; } | crontab -

but running this command gives the error:

-bash: */1 * * * * some_user python /mount/share/script.py >> /root/foo/logs/foo.cron.log 2>&1: No such file or directory

Solution

but running this command gives the error:

-bash: */1 * * * * some_user python /mount/share/script.py >> /root/foo/logs/foo.cron.log 2>&1: No such file or directory

---

The code:

(crontab -l | grep '/mount/share/script.py') || { crontab -l; '*/1 * * * * some_user python /mount/share/script.py >> /root/foo/logs/foo.cron.log 2>&1'; } | crontab -

---

will try to execute/run the:

'*/1 * * * * some_user python /mount/share/script.py >> /root/foo/logs/foo.cron.log 2>&1'

If and when grep failed.

---

Add an echo in front of it or printf, since crontab is expecting input from stdin, like what you did on your first example/code, something like:

(crontab -l | grep '/mount/share/script.py') || { crontab -l; echo '*/1 * * * * some_user python /mount/share/script.py >> /root/foo/logs/foo.cron.log 2>&1'; } | crontab -

---

Here is an alternative, which is a full blown script.

#!/usr/bin/env bash

is_in_cron='/mount/share/script.py'
cron_entry=$(crontab -l 2>&1) || exit
new_cron_entry='*/1 * * * * some_user python /mount/share/script.py >> /tmp/foo/logs/foo.cron.log 2>&1'

if [[ "$cron_entry" != *"$is_in_cron"* ]]; then
  printf '%s\n' "$cron_entry" "$new_cron_entry" | crontab -
fi

Answered By - Jetchisel

Answer Checked By - Candace Johnson (WPSolving Volunteer)