How to create a cron job using Bash automatically without the interactive editor?


Does crontab have an argument for creating cron jobs without using the editor (crontab -e). If so, What would be the code create a cronjob from a Bash script?

7/30/2017 7:13:30 AM

Accepted Answer

You can add to the crontab as follows:

#write out current crontab
crontab -l > mycron
#echo new cron into cron file
echo "00 09 * * 1-5 echo hello" >> mycron
#install new cron file
crontab mycron
rm mycron

Cron line explaination

* * * * * "command to be executed"
- - - - -
| | | | |
| | | | ----- Day of week (0 - 7) (Sunday=0 or 7)
| | | ------- Month (1 - 12)
| | --------- Day of month (1 - 31)
| ----------- Hour (0 - 23)
------------- Minute (0 - 59)

Source nixCraft.

7/9/2016 3:20:04 PM

This shorter one requires no temporary file, it is immune to multiple insertions, and it lets you change the schedule of an existing entry.

Say you have these:

croncmd="/home/me/myfunction myargs > /home/me/myfunction.log 2>&1"
cronjob="0 */15 * * * $croncmd"

To add it to the crontab, with no duplication:

( crontab -l | grep -v -F "$croncmd" ; echo "$cronjob" ) | crontab -

To remove it from the crontab whatever its current schedule:

( crontab -l | grep -v -F "$croncmd" ) | crontab -


  • grep -F matches the string literally, as we do not want to interpret it as a regular expression
  • We also ignore the time scheduling and only look for the command. This way; the schedule can be changed without the risk of adding a new line to the crontab

Thanks everybody for your help. Piecing together what I found here and elsewhere I came up with this:

The Code

command="php $INSTALL/indefero/scripts/gitcron.php"
job="0 0 * * 0 $command"
cat <(fgrep -i -v "$command" <(crontab -l)) <(echo "$job") | crontab -

I couldn't figure out how to eliminate the need for the two variables without repeating myself.

command is obviously the command I want to schedule. job takes $command and adds the scheduling data. I needed both variables separately in the line of code that does the work.


  1. Credit to duckyflip, I use this little redirect thingy (<(*command*)) to turn the output of crontab -l into input for the fgrep command.
  2. fgrep then filters out any matches of $command (-v option), case-insensitive (-i option).
  3. Again, the little redirect thingy (<(*command*)) is used to turn the result back into input for the cat command.
  4. The cat command also receives echo "$job" (self explanatory), again, through use of the redirect thingy (<(*command*)).
  5. So the filtered output from crontab -l and the simple echo "$job", combined, are piped ('|') over to crontab - to finally be written.
  6. And they all lived happily ever after!

In a nutshell:

This line of code filters out any cron jobs that match the command, then writes out the remaining cron jobs with the new one, effectively acting like an "add" or "update" function. To use this, all you have to do is swap out the values for the command and job variables.


EDIT (fixed overwriting):

cat <(crontab -l) <(echo "1 2 3 4 5") | crontab -

There have been a lot of good answers around the use of crontab, but no mention of a simpler method, such as using cron.

Using cron would take advantage of system files and directories located at /etc/crontab, /etc/cron.daily,weekly,hourly or /etc/cron.d/:

cat > /etc/cron.d/<job> << EOF
MAILTO=root HOME=/  
01 * * * * <user> <command>

In this above example, we created a file in /etc/cron.d/, provided the environment variables for the command to execute successfully, and provided the user for the command, and the command itself. This file should not be executable and the name should only contain alpha-numeric and hyphens (more details below).

To give a thorough answer though, let's look at the differences between crontab vs cron/crond:

crontab -- maintain tables for driving cron for individual users

For those who want to run the job in the context of their user on the system, using crontab may make perfect sense.

cron -- daemon to execute scheduled commands

For those who use configuration management or want to manage jobs for other users, in which case we should use cron.

A quick excerpt from the manpages gives you a few examples of what to and not to do:

/etc/crontab and the files in /etc/cron.d must be owned by root, and must not be group- or other-writable. In contrast to the spool area, the files under /etc/cron.d or the files under /etc/cron.hourly, /etc/cron.daily, /etc/cron.weekly and /etc/cron.monthly may also be symlinks, provided that both the symlink and the file it points to are owned by root. The files under /etc/cron.d do not need to be executable, while the files under /etc/cron.hourly, /etc/cron.daily, /etc/cron.weekly and /etc/cron.monthly do, as they are run by run-parts (see run-parts(8) for more information).


Managing crons in this manner is easier and more scalable from a system perspective, but will not always be the best solution.


Chances are you are automating this, and you don't want a single job added twice. In that case use:

__cron="1 2 3 4 5 /root/bin/"
cat <(crontab -l) |grep -v "${__cron}" <(echo "${__cron}")

This only works if you're using BASH. I'm not aware of the correct DASH (sh) syntax.

Update: This doesn't work if the user doesn't have a crontab yet. A more reliable way would be:

(crontab -l ; echo "1 2 3 4 5 /root/bin/") | sort - | uniq - | crontab - 

Alternatively, if your distro supports it, you could also use a separate file:

echo "1 2 3 4 5 /root/bin/" |sudo tee /etc/crond.d/backup

Found those in another SO question.


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