knowning how much time is left before cron weekly runs?
Solution 1:
The system crontab is usually defined in /etc/crontab/
. It's here that the time interval for cron.weekly, cron.daily, etc is defined. My /etc/crontab/
looks like:
SHELL=/bin/sh
PATH=/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin
# m h dom mon dow user command
17 * * * * root cd / && run-parts --report /etc/cron.hourly
25 6 * * * root test -x /usr/sbin/anacron || ( cd / && run-parts --report /etc/cron.daily )
47 6 * * 7 root test -x /usr/sbin/anacron || ( cd / && run-parts --report /etc/cron.weekly )
52 6 1 * * root test -x /usr/sbin/anacron || ( cd / && run-parts --report /etc/cron.monthly )
So my cron.weekly will run at 6:47 on the 7th day of the week.
Edit: The above information only applies to scripts in the /etc/cron.weekly directory. As pointed out by Iain, @weekly is defined to run at midnight on Sunday.
Solution 2:
Yes, by doing this on the command line:
cat /etc/crontab
Somewhere in this file, you should see a line like this:
# m h dom mon dow user command
This describes the field order for each line below it. Otherwise, it's a whitespace-separated list, where each field goes in this order:
Minute, Hour, Day of Month, Month, Day of Week (where 0 or 7 is sunday and 6 is Saturday), Username to run command as, and Command to run.
There should be a line that ends in cron.weekly. This is your weekly run. By default this runs in the early hours of Sunday morning.
Also, try running this on the command line:
man crontab
Solution 3:
I believe that @weekly
is equivalent to 0 0 * * 0
so it runs at midnight on Sunday.