Three days ago I installed the following crontab job with crontab -e:
# execute weekly
0 2 2-31 * 7 sh /home/user/folder/myscript.sh week > /home/user/.crontablog/crontab.log
It's supposed to be executed every sunday night at 2am except the 1st of the month. However it's executed every night at 2am. What's my mistake? I tried 0 instead of 7 for Sunday with the same result :/
Thank you.
Since the format of crontab is like this:
+---------------- minute (0 - 59)
| +------------- hour (0 - 23)
| | +---------- day of month (1 - 31)
| | | +------- month (1 - 12)
| | | | +---- day of week (0 - 6) (Sunday=0 or 7)
| | | | |
* * * * * command to be executed
To execute it every week on Sunday irrespective of the month you need to write it like this:
0 2 * * 7 sh /home/user/folder/myscript.sh week > /home/user/.crontablog/crontab.log
Now that you respecified your question in the comments you need to do TWO things:
as I said above, use 0 2 * * 7 to run at 2:00am every Sunday; 7 and 0 are equivalent
inside your shell script use an additional test to quietly exit on the first of the month.
That test could be as simple as
test $(date +%d) == "01" && exit
but you could of course also make it a proper if ... with more echo and verbosity.
If you want to run the script every sunday night at 2am, but only if it's not the first day in the month, then you have to use this syntax:
0 2 * * 0 /usr/bin/test $(/bin/date +\%e) -ne 1 && your_command
The test utilliy finally check if it's the first day in month and your_command is only exected if it's not.
0for Sunday, instead of7? - Steve Summit2-21for the day? Looks like it won't run if the first of the month is a Sunday. - Steve Summit0 2 * * 7to run at 2:00am every Sunday. 7 and 0 are equivalent. (And yes, I use something just like that for once-a-week jobs.) - Dirk Eddelbuettel