UNIX for Dummies Questions & Answers

Is there a way to setup a cronjob that will only run on the first monday of the month?

I must first say that I am fairly new to Unix and know very little about cronjobs, but your question intrigued me. My first question to you would be what OS are you using? The majority of my experience is with Solaris 7, so I will be speaking in terms of that.

I read the man pages on crontab and found the six fields that make up the crontab entry format:

1. Minute (0-59)
2. Hour (0-23)
3. Day of the Month (1-31)
4. Month of the Year (1-12)
5. Day of the Week (0-6) 0 being Sunday.
6. the executable string

So say you want to run your cronjob at midnight on the first Monday of ever month, it appears that the first 5 fields should look something like this:

0 0 1-7 * 1

My thinking:

0 - Is the first minute of that day
0 - Is the first hour of that day
1-7 - The first Monday has to fall within the first seven days of the month.
* - Every month
1 - Only on Mondays

Unless I am misreading the man pages, it seems like this should work. I am going to attempt to set up a cronjob and test it.

Best regards.

Sorry that I left the time out, but midnight will work fine.

0 0 1-7 * 1 command

this should execute the 'command' at Midnight on all Mondays AND first 7 days of a month.

I just need the first Monday.

You're right, molonede, "0 0 1-7 * 1 somecommand" will run command on the first 7 days of each month and on every Monday of each month. We all wish it worked the way punter5, but it doesn't. This means that you can't get what you via crontab alone.

This usual solution is something like, first write a wrapper script that will run somecommand only if the day of the month is 7 or less:

Code:

#! /usr/bin/ksh
day=$(date +%d)
if ((day <= 7)) ; then
   exec somecommand
fi
exit 1

Then run the wrapper every Monday: "0 0 * * 1 wrapper".