[SOLVED] cron expression between 07/31/2014 and 08/07/2014

zWaR · 07-22-2014, 07:39 AM

Hi all!

I'd like to construct a single cron expression which will run between July 31 and August 7 at 6:45 every day.

The following construct is not valid, because it will execute the code on August 31 as well:

Code:

45 6 31,1-7 7,8 * /myScript.sh

The problem can be solved with two cron expressions (one for July and the other for August), but can it be solved with only one expression?

Thanks for your input!

rknichols · 07-22-2014, 09:39 AM

cron can't do that directly. You would need to include a date test in the command string:

Code:

45 6 31,1-7 7,8 * [ $(date +\%j) -lt 220 ] && /myScript.sh

Just be sure to take it out of your crontab before next July, or it will run on July 1-7, 2015 and the other dates for that year as well. (Your subject did specify 2014.)

zWaR · 07-22-2014, 10:16 AM

Thanks rknichols! Good idea!

Re year, yeah sure. To make it generic one can do the following, right?

Code:

45 6 31,1-7 7,8 * [ $(date +%j) -gt 211 ] && [ $(date +%j) -lt 220 ] && /myScript.sh

rknichols · 07-22-2014, 10:53 AM

Note that I neglected to escape the "%" sign that is special in a crontab. Sorry about that. Edited to fix.

You would need to change that "220" to "221" to work right in leap years too. And, since date is a fairly complex command, you might just call it once:

Code:

45 6 31,1-7 7,8 * D=$(date +\%j); [ $D -gt 211 ] && [ $D -lt 221 ] && /myScript.sh

It looked like this was just a temporary thing since you mentioned "2014", so I didn't try to be rigorous.

zWaR · 07-22-2014, 11:02 AM

Thanks rknichols, that's really helpful!