Set Cron Job for 1st working day of every month in Shell Scripting

Question

I'm new to scripting language, can anyone please explain how to set the cron job for 1st working day?

Answer 1

You can use the following,

@monthly    

Run once a month at the morning of the first day of the month.

0 0 1 * * /home/scripts/your_script_file.sh

3rd Edit:

This will run your job at morning say 10 AM on the first weekday of the month:

# First weekday of the month

# Monday - Friday
00 10 1-3 * * [ "$(date '+\%a')" == "Mon" ] && /home/scripts/your_script_file.sh
00 10 1 * * [ "$(date '+\%a')" == "Tue" ] && /home/scripts/your_script_file.sh
00 10 1 * * [ "$(date '+\%a')" == "Wed" ] && /home/scripts/your_script_file.sh
00 10 1 * * [ "$(date '+\%a')" == "Thu" ] && /home/scripts/your_script_file.sh
00 10 1 * * [ "$(date '+\%a')" == "Fri" ] && /home/scripts/your_script_file.sh

Answer 2

First, run the date command:

$ date '+%x'
> 12/29/2014

%x tells date to display today's date in the format of the current locale. Using that exact same format, put a list of holidays in a file called holidays. For example:

$ cat holidays
> 01/01/2015
> 07/04/2015

Next, create the following shell script:

#!/bin/sh
dom=$(date '+%d') # 01-31, day of month
year=$(date '+%Y') # four-digit year
month=$(date '+%m') # two-digit month

nworkdays=0
for d in $(seq 1 $dom)
do
    today=$(date -d "$year-$month-$d" '+%x') # locale's date representation (e.g. 12/31/99)
    dow=$(date -d "$year-$month-$d" '+%u')   # day of week: 1-7 with 1=Monday, 7=Sunday
    if [ "$dow" -le 5 ]  && grep -vq "$today" /path/holidays
    then
        workday=Yes
        nworkdays=$((nworkdays+1))
    else
        workday=
    fi
done
[ "$workday" ] && [ "$nworkdays" -eq 1 ] && /path/command