CRON job to run on the last day of the month

Question

I need to create a CRON job that will run on the last day of every month. I will create it using cPanel.

Any help is appreciated. Thanks

Nephro

Answer 1

Possibly the easiest way is to simply do three separate jobs:

55 23 30 4,6,9,11        * myjob.sh
55 23 31 1,3,5,7,8,10,12 * myjob.sh
55 23 28 2               * myjob.sh

That will run on the 28th of February though, even on leap years so, if that's a problem, you'll need to find another way.

---

However, it's usually both substantially easier and correct to run the job as soon as possible on the first day of each month, with something like:

0 0 1 * * myjob.sh

and modify the script to process the previous month's data.

This removes any hassles you may encounter with figuring out which day is the last of the month, and also ensures that all data for that month is available, assuming you're processing data. Running at five minutes to midnight on the last day of the month may see you missing anything that happens between then and midnight.

This is the usual way to do it anyway, for most end-of-month jobs.

---

If you still really want to run it on the last day of the month, one option is to simply detect if tomorrow is the first (either as part of your script, or in the crontab itself).

So, something like:

55 23 28-31 * * [[ "$(date --date=tomorrow +\%d)" == "01" ]] && myjob.sh

should be a good start, assuming you have a relatively intelligent date program.

If your date program isn't quite advanced enough to give you relative dates, you can just put together a very simple program to give you tomorrow's day of the month (you don't need the full power of date), such as:

#include <stdio.h>
#include <time.h>

int main (void) {
    // Get today, somewhere around midday (no DST issues).

    time_t noonish = time (0);
    struct tm *localtm = localtime (&noonish);
    localtm->tm_hour = 12;

    // Add one day (86,400 seconds).

    noonish = mktime (localtm) + 86400;
    localtm = localtime (&noonish);

    // Output just day of month.

    printf ("%d\n", localtm->tm_mday);

    return 0;
}

and then use (assuming you've called it tomdom for "tomorrow's day of month"):

55 23 28-31 * * [[ "$(tomdom)" == "1" ]] && myjob.sh

Though you may want to consider adding error checking since both time() and mktime() can return -1 if something goes wrong. The code above, for reasons of simplicity, does not take that into account.

Rachaba

yeah it can be easier to run on every first day instead of the last day :) – Nephro

1st day of month indeed. Here's how the code will look like in PHP $date = new DateTime('2013-03-01'); $date->modify('-1 month'); $previousMonth = $date->format('Y-m'); // $previousMonth is now 2013-02. Build query to fetch products for the previous month. – Jp

Leap year feb 29 data will be lost, we need to consider that too. Thunder Rabbit answer below consider that but cron run twice in feb of leap year – Boyes

@Hari, the preferred solution would be to run on the first of the month and collect the previous month's data. Feb 29 would not be missed in that case. – Rachaba

Considering leap years: and if you don't mind it running twice: 55 23 28,29 2 * myjob.sh – Stuppy

Maybe should also add 55 23 29 2 * myjob.sh, just in case. – Muchness

You might want give credit to "Indie" as you essentially included his answer into yours. – Elva

I would think the upvotes would be enough credit, especially given the crossover is about eight lines out of sixty five :-) Just because answers may have similar information, doesn't mean they were copied, correlation is not causation and all that, I'll often go back and improve answers based on what I've learnt since the original answer, it's usually prompted by comments from people, just like this one. And, FWIW, there are four answers here that mention the tomorrow/plusoneday method so I'm not sure why you singled out just the one. – Rachaba

Answer 2

There's a slightly shorter method that can be used similar to one of the ones above. That is:

[ $(date -d +1day +%d) -eq 1 ] && echo "last day of month"

Also, the crontab entry could be update to only check on the 28th to 31st as it's pointless running it the other days of the month. Which would give you:

0 23 28-31 * * [ $(date -d +1day +%d) -eq 1 ] && myscript.sh

Carton

I couldn't get this to work as a crontab entry (something needs to be escaped I think). It worked fine in a shell script called from crontab however. FYI, the error I got was /bin/sh: -c: line 1: unexpected EOF while looking for matching ')'. – Sharecrop

This works great. In the crontab file the % must be escaped. So [ $(date -d +1day +\%d) -eq 1 ] && run_job – Dally

Nice trick indeed! But the question was tagged posix and POSIX date doesn't support "-d +1day" :-\ A more complicated (and ugly) solution would be: [ `date +\%d` -eq `cal | xargs echo | awk '{print $NF}'` ] && myscript.sh – Aphid

Answer 3

For AWS Cloudwatch cron implementation (Scheduling Lambdas, etc..) this works:

55 23 L * ? *

Running at 11:55pm on the last day of each month.

Antisemite

Answer 4

What about this one, after Wikipedia?

55 23 L * * /full/path/to/command

Hade

Well, what about it? That: "bad day-of-month errors in crontab file, can't install. Do you want to retry the same edit?" – Grassplot

Just to be clear, that wikipedia entry also mentions that "L" is non-standard. – Anastasius

Answer 5

Adapting paxdiablo's solution, I run on the 28th and 29th of February. The data from the 29th overwrites the 28th.

# min  hr  date     month          dow
  55   23  31     1,3,5,7,8,10,12   * /path/monthly_copy_data.sh
  55   23  30     4,6,9,11          * /path/monthly_copy_data.sh
  55   23  28,29  2                 * /path/monthly_copy_data.sh

Commercialism

Unless, of course, the job has some destructive aspect such as clearing out all data as it's processed :-) – Rachaba

This is actually a painless option (least technical), and you might not care that thrice in 4 years you will get the cronjob too early if you just omit the ,29. – Brumfield

@Matt: Umm, don't you mean that it will run one day too early once in four years if your crontab entry says 55   23   28    2? – Hey

@G-Man Yes, you're right, and on top of that, you have it run a second time on the 29th. – Brumfield

Answer 6

Some cron implementations support the "L" flag to represent the last day of the month.

If you're lucky to be using one of those implementations, it's as simple as:

0 55 23 L * ?

That will run at 11:55 pm on the last day of every month.

http://www.quartz-scheduler.org/documentation/quartz-1.x/tutorials/crontrigger

Engvall

Answer 7

For a safer method in a crontab based on @Indie solution (use absolute path to date + $() does not works on all crontab systems):

0 23 28-31 * * [ `/bin/date -d +1day +\%d` -eq 1 ] && myscript.sh

Celestinacelestine

Answer 8

I found out solution (On the last day of the month) like below from this site.

 0 0 0 L * ? *

CRON details:

Seconds Minutes Hours   Day Of Month    Month   Day Of Week  Year
0       0       0       L               *       ?            *

To cross verify above expression, click here which gives output like below.

2021-12-31 Fri 00:00:00
2022-01-31 Mon 00:00:00
2022-02-28 Mon 00:00:00
2022-03-31 Thu 00:00:00
2022-04-30 Sat 00:00:00

Spiculum

wait, my crontab doesn't have a seconds field... – Lighterage

this must be version dependent, I get the error: "bad day-of-month" – Lighterage

Answer 9

You could set up a cron job to run on every day of the month, and have it run a shell script like the following. This script works out whether tomorrow's day number is less than today's (i.e. if tomorrow is a new month), and then does whatever you want.

TODAY=`date +%d`
TOMORROW=`date +%d -d "1 day"`

# See if tomorrow's day is less than today's
if [ $TOMORROW -lt $TODAY ]; then
echo "This is the last day of the month"
# Do stuff...
fi

Taliped

I'm not sure what you gain by checking if it's less than today - you should just be able to check if it's the 1st. Unless the first day of the month can be the 2nd or 3rd :-) – Rachaba

Answer 10

#########################################################
# Memory Aid 
# environment    HOME=$HOME SHELL=$SHELL LOGNAME=$LOGNAME PATH=$PATH
#########################################################
#
# string         meaning
# ------         -------
# @reboot        Run once, at startup.
# @yearly        Run once a year, "0 0 1 1 *".
# @annually      (same as @yearly)
# @monthly       Run once a month, "0 0 1 * *".
# @weekly        Run once a week, "0 0 * * 0".
# @daily         Run once a day, "0 0 * * *".
# @midnight      (same as @daily)
# @hourly        Run once an hour, "0 * * * *".
#mm     hh      Mday    Mon     Dow     CMD # minute, hour, month-day month DayofW CMD
#........................................Minute of the hour
#|      .................................Hour in the day (0..23)
#|      |       .........................Day of month, 1..31 (mon,tue,wed)
#|      |       |       .................Month (1.12) Jan, Feb.. Dec
#|      |       |       |        ........day of the week 0-6  7==0
#|      |       |       |        |      |command to be executed
#V      V       V       V        V      V
*       *       28-31   *       *       [ `date -d +'1 day' +\%d` -eq 1 ] && echo "Tomorrow is the first today now is  `date`" >> ~/message
1       0       1       *       *       rm -f ~/message
*       *       28-31   *       *       [ `date -d +'1 day' +\%d` -eq 1 ] && echo "HOME=$HOME LOGNAME=$LOGNAME SHELL = $SHELL PATH=$PATH" 

Copier

Answer 11

Set up a cron job to run on the first day of the month. Then change the system's clock to be one day ahead.

Apophysis

which means the system clock will be wrong all the times. Sorry but I think this will cause more issue. -1 then. – Cardoza

Now i know how Galileo felt. – Apophysis

I think this was meant to be a joke. (At least I hope so!) It's so kludgey it could be on dailywtf.com. – Barbey

@iconoclast: I'd like to think of it as more of a koan than a joke. – Apophysis

I consider it's bad practice to change system clock, especially when you can use programming logic over data – Kristin

I think this is a really bad AND brilliant idea. Don't do this at home, kids. – Pluri

@pascalbetz Oh, people can do it at home, but they really shouldn't do it at work. – Apophysis

Run a separate server and change the time there? or Use docker to do it. May be i am taking this too far – Lynettalynette

This has a certain beauty in its naivety - I'm sure it's not serious, but there are people I've met I wouldn't put it past. – Marvelous

Answer 12

00 23 * * * [[ $(date +'%d') -eq $(cal | awk '!/^$/{ print $NF }' | tail -1) ]] && job

Check out a related question on the unix.com forum.

Irradiation

Answer 13

You can just connect all answers in one cron line and use only date command.

Just check the difference between day of the month which is today and will be tomorrow:

0 23 * * * root [ $(expr $(date +\%d -d '1 days') - $(date +\%d)  ) -le 0 ]  && echo true

If these difference is below 0 it means that we change the month and there is last day of the month.

Gunfire

Answer 14

55 23 28-31 * * echo "[ $(date -d +1day +%d) -eq 1 ] && my.sh" | /bin/bash 

Cygnus

echoing the command and piping it into bash is the best way when working with cron. Since cron is using sh and not bash. Check where your bash is using 'which bash'. On FreeBSD it is /usr/local/bin/bash, on Linux /bin/bash. – Cygnus

Answer 15

In tools like Jenkins, where usually there is no support for L nor tools similar to date, a cool trick might be setting up the timezone correctly. E.g. Pacific/Kiritimati is GMT+14:00, so if you're in Europe or in the US, this might do the trick.

TZ=Pacific/Kiritimati \n H 0 1 * * 

Result: Would last have run at Saturday, April 30, 2022 10:54:53 AM GMT; would next run at Tuesday, May 31, 2022 10:54:53 AM GMT.

Cyril

Answer 16

What about this?

edit user's .bashprofile adding:

export LAST_DAY_OF_MONTH=$(cal | awk '!/^$/{ print $NF }' | tail -1)

Then add this entry to crontab:

mm hh * * 1-7 [[ $(date +'%d') -eq $LAST_DAY_OF_MONTH ]] && /absolutepath/myscript.sh

Aggappora

Answer 17

Use the below code to run cron on the last day of the month in PHP

$commands = '30 23 '.date('t').' '.date('n').' *';

Trinitrobenzene

Answer 18

The last day of month can be 28-31 depending on what month it is (Feb, March etc). However in either of these cases, the next day is always 1st of next month. So we can use that to make sure we run some job always on the last day of a month using the code below:

0 8 28-31 * * [ "$(date +%d -d tomorrow)" = "01" ] && /your/script.sh

Strata

In Ubuntu 22.04 it seems you have to escape % symbol in cron files. So this would become 0 8 28-31 * * [ "$(date +\%d -d tomorrow)" = "01" ] && /your/script.sh – Tome

Answer 19

Not sure of other languages but in javascript it is possible.

If you need your job to be completed before first day of month node-cron will allow you to set timezone - you have to set UTC+12:00 and if job is not too long most of the world will have results before start of their month.

Distraction

Answer 20

Better way to schedule cron on every next month of 1st day

This will run the command foo at 12:00AM.

0 0 1 * * /usr/bin/foo

Callen

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – Dine

not better if you need everything to run before midnight (for instance, if you have other things like automatic upgrades and reboots scheduled for after midnight which would interfere) – Lighterage

Answer 21

Be cautious with "yesterday", "today", "1day" in the 'date' program if running between midnight and 1am, because often those really mean "24 hours" which will be two days when daylight saving time change causes a 23 hour day. I use "date -d '1am -12 hour' "

Matelda

Answer 22

If the day-of-the-month field could accept day zero that would very simply solve this problem. Eg. astronomers use day zero to express the last day of the previous month. So

00 08 00 * * /usr/local/sbin/backup

would do the job in simple and easy way.

Parisian