Cron expression for particular date

Question

I want a cron expression that represents 6th September 2010 6:00 am

Answer 1

Original question was tagged cron so this first section applies to that. See below for an updated answer for the Quartz CronTrigger tool.

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Most crontabs don't let you specify the year so you'll probably have to put that in the script itself (or a wrapper around the script/program).

You can do this with something like:

if [[ $(date +%Y) != 2010 ]] ; then
    exit
fi

The option you're looking for to run at 6am on September 6 every year is

0 6 6 9 * your_command_goes_here
| | | | |
| | | | +- any day of the week.
| | | +--- 9th month (September).
| | +----- 6th day of the month.
| +------- 6th hour of the day.
+--------- Top of the hour (minutes = 0).

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For the Quartz CronTrigger format, you'd be looking at something like:

0 0 6 6 9 ? 2010
| | | | | |   |
| | | | | |   +- 2010 only.
| | | | | +----- any day of the week.
| | | | +------- 9th month (September).
| | | +--------- 6th day of the month.
| | +----------- 6th hour of the day.
| +------------- Top of the hour (minutes = 0).
+--------------- Top of the minute (seconds = 0).

(details garnered from here).

Answer 2

For one-shot jobs the 'at' command is better suited than cron.

Answer 3

0 0 6 6 9 ? 2010
second minute hour[24 hr format] day-of-month month day-of-week year[optional]