问题
I want a cron expression that represents 6th September 2010 6:00 am
回答1:
Original question was tagged cron so this first section applies to that. See below for an updated answer for the Quartz CronTrigger tool.
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).
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).
回答2:
For one-shot jobs the 'at' command is better suited than cron.
at -f filename 06:00 09/06/2010
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回答3:
It could be only in /etc/crontab by follows:
0 6 6 9 * root test `/bin/date +%Y` == 2010 && <your command>
来源：https://stackoverflow.com/questions/3665441/cron-expression-for-particular-date