Shell Programming and Scripting

Hi,
I need subtract two date values (which are in day of the year format) and the output would give the remaining days. using the command date +"%j" i would get today's 'day of the year' i.e.,

Code:

> date +"%j"
256

Next, i need to take input of a previous date in the format 09/05/2012 and then convert it to 'day of the year format'. Could anyone please help how i could do this?
Note: I'm using HP-UX

Do you have GNU date? if yes you can refer its man page.. if not you have to rely on good old days of date logic convert them to minutes and do minus then convert back to number of days or years so...

See if you find this useful: https://www.unix.com/unix-dummies-que...n-2-dates.html

Code:

#/bin/ksh

# date format: mm/dd/yyyy

typeset -Z -R2 mm dd

y=${1##*/}
m=${1%%/*}
d=${1#*/}
d=${d%/*}
mm=$m
dd=$d
date="$mm/$dd/$y"

doy=0
f=0;
for mm in 1 2 3 4 5 6 7 8 9 10 11 12
do
   cal $mm $y | while read line
   do
      line=${line%%[a-zA-Z]*}
      [[ -n ${line:-""} ]] && {
         for dd in $line
         do
            (( doy = doy + 1 ))
            [[ "$mm/$dd/$y" = "$date" ]] && { f=1 ; break ; }
         done
         [[ "$mm/$dd/$y" = "$date" ]] && { break ; }
      }
   done
   [[ "$mm/$dd/$y" = "$date" ]] && { break ; }
done

[[ $f = 0 ]] && doy="Date not found"
echo $doy

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