Question 1

How can I create a cron job that will run on the last day of every month?

Accepted Answer

Standard cron has no 'last day of month' value. Use three lines: `0 0 30 4,6,9,11 * command` (30-day months), `0 0 31 1,3,5,7,8,10,12 * command` (31-day months), `0 0 28,29 2 * command` (February, leap-year safe). On schedulers that support it (Quartz, AWS, robfig/cron) use the L modifier: `0 0 L * *`.

Question 2

Can I run a cronjob every three days?

Accepted Answer

Use `0 0 1-30/3 * * command`. The day-of-month `1-30/3` expands to 1, 4, 7, 10, …, 28 — ten runs per month at every 3rd day starting from the 1st. The cadence resets at each month boundary, so February's first run lands on the 1st even if the previous month's last was on the 28th. For strict 72-hour intervals, use a job runner instead.

Question 3

How do I set a cron job to run twice a week (bi-weekly)?

Accepted Answer

Use `0 0 * * 1,4 command` for twice-weekly (Monday and Thursday) or `0 0 * * MON,THU command` with name aliases. True bi-weekly (every other Monday) isn't native to cron — wrap a weekly cron with a script that exits on odd/even ISO weeks: `0 9 * * 1 [ $(($(date +%V) % 2)) -eq 0 ] && /path/to/script.sh`.

Question 4

I am trying to create a Cron expression that runs every day at 8:00 AM and 3:30 PM. I understand how to create an expression that runs once a day, but not at multiple set times.

Accepted Answer

Use two cron lines, one per fixed time: `0 8 * * * command` for 08:00 daily and `30 15 * * * command` for 15:30 daily. Standard cron only allows a list per field, not paired hour:minute combinations, so different minutes-per-hour always require multiple cron entries.

How can I create a cron job that will run on the last day of every month?

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