I need to get yesterdays date in the format yyyymmdd
I can get today's date simply enough - 20031112

Is there any way to substract 1 from this easily enough in korn shell script?

It has to be korn shell and not perl

Hi,

You can use the script below, it will do more than you asked, but you can easly modify it... it was written by Tapani Tarvainen.

#! /usr/bin/ksh
# Get yesterday's date in YYYY-MM-DD format.
# With argument N in range 1..28 gets date N days before.
# Tapani Tarvainen January 2002
# This code is in the public domain.

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

# Subtract offset from day, if it goes below one use 'cal'
# to determine the number of days in the previous month.
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
print $month/$day/${year#??}

Thanks - I have found a way myself now...


datestamp=`date '+%Y%m%d'`
yest=$((datestamp -1))



Cheers!!

That won't work on the first of the month.

Here is a script that can do date calculations without the use of external programs.

Hi- Thanks - to be honest I hadnt thought of the first of the month

I am using your script as follows:
./datecalc -a 2003 11 12 - 1

and it returns 2003 11 11

Is there a way that it can return the value without the spaces?

./datecalc -a 2003 11 12 - 1 | read y m d ; echo ${y}${m}${d}

# ./datecalc -a 2003 11 12 - 1 | read y m d ; echo ${y}${m}${d}
y: Undefined variable


any ideas?