Shell Programming and Scripting

Have you looked at "TZ" like in this post:
https://www.unix.com/unix-dummies-que...days-date.html

I knew I'd seen this someplace, as per the guidelines, remember to do a search of the forums first.
This should answer your question:
https://www.unix.com/answers-frequent...rithmetic.html

I hope its not a late reply mate, i have used such a good script as a part of back up script and it works fine, i hope its usefull for u:

#! /usr/bin/ksh

OFFSET=${1:-1}

case $OFFSET in
*[!0-9]* | ???* | 3? | 29) print -u2 "Invalid input" ; exit 1;;
esac

eval `date "+day=%d; month=%m; year=%Y`
typeset -Z2 day month
typeset -Z4 year
day=$((day - OFFSET))
if (( day <= 0 )) ;then
month=$((month - 1))
if (( month == 0 )) ;then
year=$((year - 1))
month=12
fi
set -A days `cal $month $year`
xday=${days[$(( ${#days[*]}-1 ))]}
day=$((xday + day))
fi

print $year$month$day



Regards

Try this

now=`date -d "15 Aug 2009" "+%s"
perl -e 'print scalar localtime('"$now"'-24*3600), "\n"'


Regards,

Ranjith

By setting timezone are GMT+24 can give you previous day

$ date
Thu Aug 20 13:59:27 BST 2009
$ TZ=GMT+24
$ date
Wed Aug 19 12:59:37 GMT 2009

setting TZ back to two days
$ TZ=GMT+48
$ date
Tue Aug 18 13:01:22 GMT 2009

Hi thanks for your reply.
I want to give an input as yyyy-mm-dd