UNIX for Dummies Questions & Answers

#!/bin/bash
#
#

# check the command line
if [ $# -ne 1 ]
then
    echo "Usage: ${0##*/} <year and day of year> (i.e. 2013159)"
    exit 1
fi

# get the given year
yr=$(echo $1 | cut -c1-4)

# get the given day
dy=$(echo $1 | cut -c5-7)

# perform the conversion using the date command
output=$(date -d "${yr}-01-01 +$(( ${dy} - 1 ))days" +%Y%m%d)

# display results
echo "$output"

# done
exit 0

day2date 2013159
20130608

MONTHS=(31 28 31 30 31 30 31 31 30 31 30 31)
DATE=2013159
YEAR=${DATE:0:4}
DAY=${DATE:4}
((MONTHS[1]+=YEAR%4==0))
while ((DAY > (sum+=MONTHS[i++]))) ; do : ; done; printf "%d%02d%02d\n" $YEAR $i $((DAY+MONTHS[$i-1]-sum)) 
20130608

dconv
       Convert DATE/TIMEs between calendrical systems.  If DATE/TIME is
       omitted date/times are read from stdin.

dconv -i '%Y%D' 2013159
=> 2013-06-08