run-if-today

#!/usr/bin/env bash

# run-if-today
# Run a cron task the first, nth or last weekday of the month.
#
# Homepage:
# https://github.com/xr09/cron-last-sunday
#
# Released under the MIT license.

# Examples:
#
# 1st Monday	--> run-if-today 1 "Mon"
# 2nd Friday	--> run-if-today 2 "Fri"
# last Sunday	--> run-if-today [l|L] "Sun"

# Date ranges:
#
#	Week	Dates
#	1	1-7
#	2	8-14
#	3	15-21
#	4	22-28
#	5	29-lastday	(you probably need [L]ast week instead)
#	L	(lastday-6)-lastday


ARRAY=( $(LC_TIME=en_US.UTF-8 date "+%Y %m %d %a") )
YEAR=${ARRAY[0]}
MONTH=${ARRAY[1]}
TODAY=${ARRAY[2]}
WEEK_DAY=${ARRAY[3]}

# If the day is provided but it is not that day of the week, don run all the checks
if [ ! -z "$2" ] && [ $WEEK_DAY != ${2:0:3} ]
then
	exit 1
fi

# Ok, it's that day, let's run some fancy code.

# Which week in the month are we dealing with?
case "${1:0:1}" in
	[1-5])
		# Plain old nth day code.
		MIN_DAY=$[ 1 + $[ 7 * $[ ${1:0:1} - 1 ] ] ]
		MAX_DAY=$[ $MIN_DAY + 7 ]

		if [ $TODAY -ge $MIN_DAY ] && [ $TODAY -lt $MAX_DAY ]
		then
			# Green light
			exit 0
		fi
		;;
	"l" | "L")
		# April, June, September and November have 30 days.
		# All others have 31 except February
		case $MONTH in
			"02")
				# First we assume it's a normal February.
				LAST_DAY=28

				# Then we test if this is a leap year.
				if [ $(date -d "$YEAR-02-29" > /dev/null 2>&1) ]
				then
					# It's a leap year
					LAST_DAY=29
				fi
				;;
			"04" | "06" | "09" | "11")
				LAST_DAY=30
				;;
			*)
				LAST_DAY=31
				;;
		esac

		# 7 days from the end of the month.
		START_OF_LAST_WEEK=$[ $LAST_DAY - 6 ]

		if [ $TODAY -ge $START_OF_LAST_WEEK ] && [ $TODAY -le $LAST_DAY ]
		then
			# This is last week of the month.
			exit 0
		fi
		;;
	*)
		echo "$0: Unknown argument. Was expecting: 1, 2, 3, 4, 5, l or L"
		exit 2
		;;
esac

# Not today
exit 1