The second Saturday of the month falls on one (and only one) of the dates from the 8th to the 14th inclusive. Likewise, the fourth Saturday falls on one date between the 22nd and the 28th inclusive.
You may think that you could use the day of week field to limit it to Saturdays (the 6
in the line below):
0 1 8-14,22-28 * 6 /path/to/myscript
Unfortunately, the day-of-month and day-of-week is an OR proposition where either will cause the job to run, as per the manpage:
Commands are executed by cron
when the minute, hour, and month of year fields match the current time, and when at least one of the two day fields (day of month, or day of week) match the current time.
The day of a command's execution can be specified by two fields - day of month, and day of week. If both fields are restricted (i.e., aren't *), the command will be run when either field matches the current time. For example, 30 4 1,15 * 5
would cause a command to be run at 4:30 am on the 1st and 15th of each month, plus every Friday.
Hence you need to set up your cron
job to run on every one of those days and immediately exit if it's not Saturday.
The cron
entry is thus:
0 1 8-14,22-28 * * /path/to/myscript
(to run at 1am on each possible day).
Then, at the top of /path/to/myscript
, put:
# Exit unless Saturday
if [[ $(date +%u) -ne 6 ]] ; then
exit
fi
And, if you can't modify the script (e.g., if it's a program), simply write a script containing only that check and a call to that program, and run that from cron
instead.
You can also put the test for Saturday into the crontab
file itself to keep the scheduling data all in one place:
0 1 8-14,22-28 * * [ `date +\%u` = 6 ] && /path/to/myscript