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

You can add to the crontab as follows:

#write out current crontabcrontab -l > mycron#echo new cron into cron fileecho "00 09 * * 1-5 echo hello" >> mycron#install new cron filecrontab mycronrm 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.

You may be able to do it on-the-fly

crontab -l | { cat; echo "0 0 0 0 0 some entry"; } | crontab -

crontab -l lists the current crontab jobs, cat prints it, echo prints the new command and crontab - adds all the printed stuff into the crontab file. You can see the effect by doing a new crontab -l.

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 -

Notes:

• 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