Debian – Cron allowed special character “L” for “Last day of the month” on Debian

crondebian

I'd like to know if L is one of the allowed special characters on Debian's cron implementation? I'm trying to set up a cron to run on the last day of every month.

From the cron entry on wikipedia:

'L' stands for "last". When used in the day-of-week field, it allows
you to specify constructs such as "the last Friday" ("5L") of a given
month. In the day-of-month field, it specifies the last day of the
month.

Note: L is a non-standard character and exists only in some cron
implementations (Quartz java scheduler)

If not, how would you go about setting a cron to run on the last day of every month? Would you recommend 3 different entries like this solution on stackoverflow?

Best Answer

Cron entries on Debian are described in the crontab man page (man 5 crontab). Debian uses Vixie's cron, and his man page says:

   The crontab syntax does not make it possible  to  define  all  possible
   periods  one could image off. For example, it is not straightforward to
   define the last weekday of a month. If a task needs to be run in a spe-
   cific  period of time that cannot be defined in the crontab syntaxs the
   best approach would be to have the program itself check  the  date  and
   time  information and continue execution only if the period matches the
   desired one.

   If the program itself cannot do the checks then a wrapper script  would
   be required. Useful tools that could be used for date analysis are ncal
   or calendar For example, to run a program the last  Saturday  of  every
   month you could use the following wrapper code:

   0 4 * * Sat   [ "$(date +%e)" = "`ncal | grep $(date +%a | sed  -e 's/.$//') 
     | sed -e 's/^.*\s\([0-9]\+\)\s*$/\1/'`" ] && echo "Last Saturday" &&
     program_to_run

So working along those lines:

   0 0 * * * perl -MTime::Local -e 
       'exit 1 if (((localtime(time()+60*60*24))[3]) < 2);' || program_to_run

Related Solutions

How to tell cron to run a command every other day (odd/even)

The syntax you tried is actually ambiguous. Depending on how many days are in the month, some months it will run on odd days and some on even. This is because the way it is calculated takes the total number of possibilities and divides them up. You can override this strage-ish behavior by manually specifying the day range and using either an odd or even number of days. Since even day scripts would never run on the 31st day of longer months, you don't lose anything using 30 days as the base for even-days, and by specifying specifically to divide it up as if there were 31 days you can force odd-day execution.

The syntax would look like this:

# Will only run on odd days:
0 0 1-31/2 * * command

# Will only run on even days:
0 0 2-30/2 * * command

Your concern about months not having the same number of days is not important here because no months have MORE days than this, and for poor February, the date range just won't ever match the last day or two, but it will do no harm having it listed.

The only 'gotcha' for this approach is that if you are on an odd day cycle, following months with 31 days your command will also run on the first of the month. Likewise if you are forcing an even cycle, each leap year will cause one three-day cycle at the end of February. You cannot really get around the fact that any regular pattern of "every other day" is not always going to fall on even or odd days in every month and any way you force this you will either have an extra run or be missing a run between months with mismatched day counts.

How to format these two complex cron statements

To have something execute only on the second Monday of a month the day of week value needs to be 1 and the day of month value has to be 8-14, the hour has to be 2,6,10,14,18,22 and the minute 0. However as dhag correctly commented and provided a solution for, when you specify both the day of week and the day of month (i.e. not as *), the program is executed when either matches. Therefore you have to test explicitly for either one, and the day of week is easier:

 0 2,6,10,14,18,22 8-14 * * test $(date +\%u) -eq 1 && /opt/bin/cleanup.sh

The final 1 determines the Monday and the range of day of month (8-14) picks it only when it is the second Monday.

The third day of every month ending in an "r" at 18:15:

 15 18 3 september,october,november,december * /opt/bin/verrrrrrrry.sh

(at least on Vixie cron you can use the names of the months. If yours does not support that you can replace that with 9-12)

Best Answer

Related Solutions

How to tell cron to run a command every other day (odd/even)

How to format these two complex cron statements

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