Shell – Schedule the last day of every month

crondateshell-scripttest

I read from an instruction to schedule a script on the last day of the month:

Note:
The astute reader might be wondering just how you would be able to set a command to execute on the last day of every month because you can’t set the dayofmonth value to cover every month. This problem has plagued Linux and Unix programmers, and has spawned quite a few different solutions. A common method is to add an if-then statement that uses the date command to check if tomorrow’s date is 01:
00 12 * * * if [`date +%d -d tomorrow` = 01 ] ; then ; command1
This checks every day at 12 noon to see if it's the last day of the month, and if so, cron runs the command.

How does [`date +%d -d tomorrow` = 01 ] work?
Is it correct to state then; command1?

Best Answer

Abstract

The correct code should be:

#!/bin/sh
[ "$#" -eq 0 ] && echo "Usage: $0 command [args]" && exit 1
[ "$(date -d tomorrow +'%d')" = 01 ] || exit 0
exec "$@"

Call this script end_of_month.sh and the call in cron is simply:

00 12 28-31 * * /path/to/script/end_of_month.sh command

That would run the script end_of_month (which internally will check that the day is the last day of the month) only on the days 28, 29, 30 and 31. There is no need to check for end of month on any other day.

Old post.

Yes, that is what it says, but that is wrong/incomplete:

Missing a closing fi.
Needs space between [ and the following `.
It is strongly recommended to use $(…) instead of `…`.
It is important that you use quotes around expansions like "$(…)"
There is an additional ; after then

How do I know? (well, by experience ☺ ) but you can try Shellcheck. Paste the code from the book (after the asterisks) and it will show you the errors listed above plus a "missing shebang". An script without any errors in Shellcheck is this:

~~#!/bin/sh if [ "$(date +%d -d tomorrow)" = 01 ] ; then script.sh; fi~~

That site works because what was written is "shell code". That is a syntax that works in many shells.

Some issues that shellcheck doesn't mention are:

It is assuming that the date command is the GNU date version. The one with a -d option that accepts tomorrow as a value (busybox has a -d option but doesn't understand tomorrow and BSD has a -d option but is not related to "display" of time).
It is better to set the format after all the options date -d tomorrow +'%d'.
The cron start time is always in local time, that may make one job start 1 hour earlier of later than an exact day count if the DST (daylight saving time) got set or unset.

What we got done is a shell script which could be called with cron. We can further modify the script to accept arguments of the program or command to execute, like this (finally, the correct code):

#!/bin/sh
[ "$#" -eq 0 ] && echo "Usage: $0 command [args]" && exit 1
[ "$(date -d tomorrow +'%d')" = 01 ] || exit 0
exec "$@"

Call this script end_of_month.sh and the call in cron is simply:

00 12 28-31 * * /path/to/script/end_of_month.sh command

Make sure the correct path is included. The PATH inside cron will not (not likely) be the same as the user PATH.

Note that there is one end of month script tested (as indicated below) that could call many other utilities or scripts.

This will also avoid the additional problem that cron generates with the full command line:

Cron splits the command line on any % even if quoted either with ' or " (only a \ works here). That is a common way in which cron jobs fail.

You can test if end_of_month.sh script works correctly on some date (without waiting to the end of the month to discover it doesn't work) by testing it with faketime:

$ faketime 2018/10/31 ./end_of_month echo "Command will be executed...."
Command will be executed....

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.

Bash – In linux, would it be possible to run a script every day 3 minutes later than the previous day

You can use the at command from within your script (or a wrapper). Give it the time to run the next iteration.

echo '/dir/scriptname' | at 'now + 1443 minutes'

Put that line as near as possible to the beginning of the script to reduce drift.