Run a cron job on the first Monday of every month?

Question

I'd like to run a job from cron at 8.30 on the first Monday of every month. The cron Wikipedia page says

While normally the job is executed when the time/date specification fields all match the current time and date, there is one exception: if both "day of month" and "day of week" are restricted (not "*"), then either the "day of month" field (3) or the "day of week" field (5) must match the current day.

(my emphasis)

Does that mean I can't do the first Monday of the month, I can only do the first (or whatever) day of the month? I can't think of a way round it.

slhck - that sounds like a good solution - would you mind expanding it into an answer with the full code? then i'll mark it as correct :) — Max Williams, May 28 '12 at 08:51
Similar answer in https://superuser.com/questions/1042805/how-to-set-cron-expression-to-run-first-weekday-mon-fri-and-first-weekend-sat — Michael Freidgeim, May 01 '21 at 21:15

score 57 · Accepted Answer · edited Jan 09 '21 at 10:08

57

You can put the condition into the actual crontab command (generic way):

[ "$(date '+%u')" = "1" ] && echo "It's Monday"

if your locale is EN/US, you can also compare strings (initial answer):

[ "$(date '+%a')" = "Mon" ] && echo "It's Monday"

Now, if this condition is true on one of the first seven days in a month, you have its first Monday. Note that in the crontab, the percent-syntax needs to be escaped though (generic way):

0   12  1-7 *   *   [ "$(date '+\%u')" = "1" ] && echo "It's Monday"

if your locale is EN/US, you can also compare strings (initial answer):

0   12  1-7 *   *   [ "$(date '+\%a')" = "Mon" ] && echo "It's Monday"

Replace the echo command with the actual command you want to run. I found a similar approach too.

edited Jan 09 '21 at 10:08

4wk_

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answered May 28 '12 at 09:02

slhck

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actually, slhck, i spoke too soon - this isn't working in my crontab: it seems to have a problem with the date test for some reason. This works in the command line `[ "$(date '+%a')" = "Mon" ] && echo "It's Monday"` as does `[ $(date '+%a') = "Mon" ] && echo "It's Monday"` But, in the crontab it seems to be failing at this test. I can't get at any useful feedback from cron either: i'm trying this `[ "$(date '+%a')" = "Tue" ] >> /home/deploy/cron.log 2>&1` to capture any error output into a log but that's not working either :/ – Max Williams May 29 '12 at 08:53
Hm. Can you check whether the `date` output is correct? For example, just let a `cron` command run that outputs this to a file: `date '+%a' > ~/datetest 2>&1` – then check if that resulted in the right string to compare against. – slhck May 29 '12 at 08:58
3

I found the problem - the percent in `+%a` needed escaping when it was in the crontab, ie the line needed to be this: `[ "$(date '+\%a')" = "Mon" ] && echo "It's Monday"`. If you could update your answer with that it would be helpful to future readers :) Thanks again - max – Max Williams May 29 '12 at 09:33
Oh, now I see why. Glad it works now, thanks for trying it out! – slhck May 29 '12 at 09:35
1

Beware that for alpine's CRON implementation, the percent sign **needs not** to be escaped. – helvete Feb 17 '21 at 14:55

score 28 · Answer 2 · answered Oct 24 '18 at 23:58

28

I have a computer with locale on Spanish, so, this approach isn't working for me because mon changes to lun

Other languages would fail as well, so, I did a slight variation on the accepted answer that takes out the language barrier:

 0 9 1-7 * *   [ "$(date '+\%u')" = "1" ] && echo "¡Es lunes!"

answered Oct 24 '18 at 23:58

cumanacr

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This should be the accepted answer. I have a Dutch locale which also uses different names for weekdays. And using numbers also feels cleaner than comparing strings to determine the day of the week. – 0ne_Up Nov 28 '18 at 10:54

score 11 · Answer 3 · edited Aug 06 '15 at 11:41

11

I find it easier when there's no need to handle day numbers.

Run First Monday of month:

0 2 * * 1 [ `date '+\%m'` == `date '+\%m' -d "1 week ago"` ] || /path/to/command

i.e. if the month 1 week ago is not the same as the current month then we are on the 1st day 1 (= Monday) of the month.

Similarly, for the Third Friday

0 2 * * 6 [ `date '+\%m'` == `date '+\%m' -d "3 weeks ago"` ] || /path/to/command

i.e. if the month 3 weeks ago is different to current month then we are on the 3rd day 6 (= Friday) of the month

edited Aug 06 '15 at 11:41

Joel

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answered Aug 06 '15 at 08:12

joel

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Day 6 of week is Saturday not Friday. – Robert Jun 05 '19 at 09:23

score 7 · Answer 4 · answered Aug 09 '18 at 11:14

7

Since I interpret my cron statements using php and js, I can't use bash. Finally I found that it is in fact possible with just cron:

0 30 8 * 1/1 MON#1

Hope this helps someone else. Regardless, I wish you all a beautiful day. :-)

answered Aug 09 '18 at 11:14

kramer65

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This is not cron, it's possibly quartz – sivann Nov 29 '19 at 07:16
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According to https://crontab.cronhub.io/ it is cron. Can also try 0 0 8 ? * 1#1 * – dbinott Oct 26 '22 at 15:12

score 5 · Answer 5 · edited Jan 18 '13 at 09:40

5

I have scheduled a job to run on the 4th Monday of every month at 4:00 PM as follows:

0 16 22-28 * Mon [ "$(date '+\%a')" == "Mon" ] && touch /home/me/FourthMonOfMonth.txt

edited Jan 18 '13 at 09:40

Indrek

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answered Jan 18 '13 at 08:06

Abhijit Mapgaonkar

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Would have the same effect as 0 16 22-28 * * , but would run four times a month more. – user3132194 Mar 07 '23 at 05:53

score 3 · Answer 6 · answered Sep 23 '22 at 10:23

There is a hacky way to do this with a classic (Vixie, Debian) cron:

30 8 */100,1-7 * MON

The day-of-month field starts with a star (*), and so cron considers it "unrestricted" and uses the AND logic between the day-of-month and the day-of-week fields.

*/100 means "every 100 days starting from date 1". Since there are no months with more than 100 days, */100,1-7 effectively means "on dates 1 to 7".

Here's my article with more details: Schedule Cronjob for the First Monday of Every Month, the Funky Way

score 2 · Answer 7 · edited Nov 29 '18 at 19:28

2

I recommend to use

"$(/bin/date '+%\w')" = "1"

instead of

"$(date '+\%a')" = "Mon"

to avoid locale problem.

edited Nov 29 '18 at 19:28

Tomasz Jakub Rup

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answered Nov 29 '18 at 18:28

Manus Freedom

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score 1 · Answer 8 · answered May 25 '12 at 15:08

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As far as I know it is NOT possible using only crontab, however one can use a wrapper function to pick the correct day from a "first seven days of month" contab entry; see this from entry.

The wrapper script would be

#! /usr/bin/ksh
day=$(date +%d)
if ((day <= 7)) ; then
   exec somecommand
fi
exit 1

and you would need to run it (assuming it is called wrapper.sh and globally available) using the crontab entry

0 0 * * 1 wrapper.sh

answered May 25 '12 at 15:08

elemakil

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thanks elemaki. I thought about a similar solution using some extra code inside the task that gets run, getting it to test whether it's the first monday of the month inside the script, and then calling it every monday in crontab. It's a ruby script so it's easy to test the day of th week. But i was hoping there'd be a crontab only way. – Max Williams May 25 '12 at 15:56

score 1 · Answer 9 · answered Oct 02 '15 at 15:25

1

On Solaris 10 I had to format the condition as follows:

[ `date +\%a` = "Sat" ] && echo "It's Saturday"

answered Oct 02 '15 at 15:25

Will

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score 1 · Answer 10 · answered Oct 21 '15 at 13:24

1

You can try running cronjob for first seven days of month and let it execute only on Monday.

30 8 * * 1 [`date +\%d` -le 07] && <job>

Above should work for you.

answered Oct 21 '15 at 13:24

simer

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+1 more success hit ration then to run for the first week and check for Monday. :-) – xmedeko Mar 22 '17 at 07:28

Chris Hubbard · Answer 11 · 2020-03-08T23:39:42.397

0

This usage should be most universal and avoids the locale issue.

[ `/bin/date +\%u` -eq 1 ]

the first Monday of the month at 6am would look like this in /etc/crontab

00 6 1-7 * *    root    [ `/bin/date +\%u` -eq 1 ] && /run/yourjob.sh

edited Mar 08 '20 at 23:39

answered Mar 08 '20 at 22:45

Chris Hubbard

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SRGDVD · Answer 12 · 2022-11-17T18:38:36.550

0

crontab 30 8 */27 * 1

At 08:30 AM, [ones] every 27 days, and on Monday (see crontab expression generator)

Day of the month is unencumbered (is *), so the either logic/exception does not apply.

edited Nov 17 '22 at 18:38

answered Nov 16 '22 at 22:18

SRGDVD

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Chirag Pansheriya · Answer 13 · 2023-01-17T13:21:13.037

0

The day of month field here is */100,1-7, meaning “every 100 days starting from date 1, and also on dates 1-7”. Since there are no months with 100+ days, this again is a trick to say “on dates 1 to 7” but with a leading star. Because of the star, cron will run the command on dates 1 to 7 that are also Monday.

0 22 */100,1-7 * 2

On the cron guru example below, you can see the result and the next run date.

https://crontab.guru/#0_0_/100,1-7__MON

edited Jan 17 '23 at 13:21

answered Jan 07 '23 at 20:47

Chirag Pansheriya

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As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Jan 07 '23 at 21:02

score 0 · Answer 14 · edited Feb 13 '23 at 17:00

0

I just ran into this and here is what I came up with:

0 11 * * 1#1 - Run at 11 am on the first Monday of every month
0 11 * * 1#2 - Run at 11 am on the second Monday of every month
0 11 * * 1#3 - Run at 11 am on the third Monday of every month

If you wanted to run on Tuesday instead:

0 11 * * 2#1 - Run at 11 am on the first Tuesday of every month
0 11 * * 2#2 - Run at 11 am on the second Tuesday of every month
0 11 * * 2#3 - Run at 11 am on the third Tuesday of every month

edited Feb 13 '23 at 17:00

Dominique

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answered Feb 13 '23 at 15:40

ErnieAndBert

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Is it some cron dialect? What sort of cron is it for? This fails check in vixie-cron. – user3132194 Mar 07 '23 at 06:17

score 0 · Answer 15 · answered Mar 04 '14 at 15:40

I made a general solution for this kind of problems, it works for first, second, third..... last weekday of the month.

You can use it like this:

30 06 * * Mon run-if-today 1 "Mon" && echo "First Monday"
30 06 * * Thu run-if-today 3 "Thu" && echo "Third Thursday"
30 06 * * Sun run-if-today L "Sun" && echo "Last Sunday"

The script run-if-today check for both the weekday and the desired week date range, if both match then it returns 0, otherwise 1.

Check the code here. https://github.com/xr09/cron-last-sunday

score -1 · Answer 16 · answered Apr 20 '21 at 02:54

-1

I needed the same result, but wanted to use pure cron. I hope this improves on the accepted answer.

For me I needed it to run the First Monday of each month at noon.

So this should run: At Noon, in the first 7 days of the month, on a Monday: 0 12 1-7 * 1

https://crontab.guru/#0_12_1-7_*_1

answered Apr 20 '21 at 02:54

Paul Perrick

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from the guru: `“At 12:00 on every day-of-month from 1 through 7 and on Monday.”` it will run every day of the week AND on Monday. – Art3mix Sep 24 '21 at 07:24

score -1 · Answer 17 · answered Jul 08 '14 at 13:26

-1

I believe this solves the problem more elegantly:

30 8 1-7 * 1 /run/your/job.sh

answered Jul 08 '14 at 13:26

symcbean

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Doh! : "If both fields are restricted (ie, aren't *), the command will be run when either field matches the current time." – symcbean Jul 10 '14 at 08:35
Also on February (the one that ends 28th) it will not be run every 4 years. – igraczech Oct 05 '16 at 11:00
"either" means any of them, not "both". this will run every day for the 1st 7 days of the month. – minusf Aug 30 '18 at 14:42
1

This row would run a command on the first to seventh of each month, as well as on every Monday – Tomasz Jakub Rup Nov 30 '18 at 05:23

score -2 · Answer 18 · answered Sep 04 '18 at 09:01

-2

0 9 1-7 * 1 *

This will work for every monday of every month.

answered Sep 04 '18 at 09:01

SIBHI S

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An answer has been accepted, is your answer better, different or useful on a newer version, etc. – mic84 Sep 04 '18 at 09:32

Run a cron job on the first Monday of every month?

18 Answers18

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