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Cron expression for first two mondays of every month
I need a cron expression that would run a report on first two mondays of every month.What would be the expression?Thanks in advance
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Not sure how flexible the cron syntax is within Splunk so mabe you could try with:
0 8 1-14 * 1
Which reads as: At 08:00 on every day-of-month from 1 through 14 that is also a Monday.
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Thanks Javiergn but it's not the requirement I want.Canwe make any changes at the last number ie 1 which shows Monday?
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What's your actual requirement?
You can make any changes you like, syntax is as follows:
* * * * *
- - - - -
| | | | |
| | | | +----- day of week (0 - 6) (Sunday=0)
| | | +------- month (1 - 12)
| | +--------- day of month (1 - 31)
| +----------- hour (0 - 23)
+------------- min (0 - 59)
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You might be correct.I took 1-14 as dates not as day.Thanks.And sorry also my mistake
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No worries. Please don't forget to confirm the answer once you have verified it fully works for you.
I'm still not 100% sure anymore given what c_krishna_guturi mentioned above so you are going to have to test and let us know.
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The expression will give the alerts at 08:00 on every day-of-month from 1 through 14 and on Monday
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You might be right. Don't have my lab to test this but I'm finding contradictory answers online:
See for instance:
VS
http://corntab.com/?c=0_8_1-14/7_*_1
I don't have my lab to test this at the moment sorry
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