UNIX for Dummies Questions & Answers

Hello,
I know this has probably been discussed million times before but I can't find anything. At my old job someone made a shell script function where you pass a parameter to the function, like 36 and it would return what date was 36 days ago. It took into account leap years. It would make it so the Month, Day, and Year were in different variables so you could build it like you want it. I found out about date -d "2 days ago" which work great, but you can't substitue the number with a variable, or can you. If so could you post your code snippett. I was thinking you could something like this:
DATE=`date -d "${DAYAGO} days ago"
But this doesn't seem to work.
But this will work

DAYAGO=$1
RUNME=`date -d "3 days ago"`
echo ${RUNME}

Though DAYAGO is not being used, anyone know how I can make it work with the variable DAYAGO.

Thanks for any help.
bill

Try...

Code:

#!/usr/bin/ksh

function days_ago {
  DAYAGO=$1
  set -- $(date -d "$DAYAGO days ago" "+%Y %m %d")
  YYYY=$1 ; MM=$2 ; DD=$3
  shift $#
}

days_ago 3
echo $YYYY-$MM-$DD

Thanks, that works great.