Quote:

Originally Posted by Franklin52 A sample to calculate the age with a given date, adapt it to read the date from a file or if you have a different date format. Code: #!/bin/sh echo -n "Enter the birthdate (mm-dd-yyyy): "

The -n option to echo is not standard.

read bdate

bmonth=${bdate:0:2}

That is not sh syntax; it is limited to bash and ksh.

bday=${bdate:3:2}
byear=${bdate:6:4}

cdate=`date +%m-%d-%Y`

cmonth=${cdate:0:2}
cday=${cdate:3:2}
cyear=${cdate:6:4}

Simpler (and POSIX compliant) is:

Code:

eval "$( date "+cyear=%Y cmonth=%m cday=%d" )"

if [[ "$cmonth" -lt "$bmonth" ]] || [[ "$cmonth" -eq "$bmonth" &&

[[ ... ]] is not sh syntax. Use [ .. ].

"$cday" -lt "$bday" ]]
then
  let age=cyear-byear-1

let is not standard. Use POSIX arithmetic:

Code:

age=$(( $cyear - $byear - 1 ))

else
  let age=cyear-byear
fi

echo "Age = "$age

This will run in any POSIX shell:

Code:

## Accept input on command line, or read interactively if none
case $# in
  3) bday=$1; bmonth=$2; byear=$3 ;;
  *) printf "Enter the birthdate (day month year): "
     read bday bmonth byear
     ;;
esac

## Ensure 2 digits for $bmonth and $bday
case $bmonth in ?) bmonth=0$bmonth;; esac
case $bday in ?) bday=0$bday;; esac

eval "$( date "+cyear=%Y cmonth=%m cday=%d" )"

[ $bmonth$bday -lt $cmonth$cday ] &&
 age=$(( $cyear - $byear )) ||
  age=$(( $cyear - $byear - 1 ))

echo "Age = $age"

Quote:

Originally Posted by fpmurphy ] Afraid it is supported by IEEE 1003.1:2004.

No, It is neither supported nor required. In fact, it's behaviour is not even specified.

A POSIX shell is not required to support it, and if it does, it could do anything at all, including erase the contents of you hard drive.