UNIX for Dummies Questions & Answers

 date 1 = 03-02-2016
 date 2 = 15-11-2015

awk -vdate1="03-02-2016" -vdate2="15-11-2015" 'BEGIN{split(date1, A,"-");split(date2, B,"-");year_diff=A[3]-B[3];if(year_diff){months_diff=A[2] + 12 * year_diff - B[2] + 1;} else {months_diff=A[2]>B[2]?A[2]-B[2]+1:B[2]-A[2]+1};print months_diff}'

awk -vdate1="03-02-2015" -vdate2="15-11-2015" 'BEGIN{split(date1, A,"-");split(date2, B,"-");year_diff=A[3]-B[3];if(year_diff){months_diff=A[2] + 12 * year_diff - B[2] + 1;} else {months_diff=A[2]>B[2]?A[2]-B[2]+1:B[2]-A[2]+1};print months_diff}'

awk -vdate1="03-02-2017" -vdate2="15-11-2015" 'BEGIN{split(date1, A,"-");split(date2, B,"-");year_diff=A[3]-B[3];if(year_diff){months_diff=A[2] + 12 * year_diff - B[2] + 1;} else {months_diff=A[2]>B[2]?A[2]-B[2]+1:B[2]-A[2]+1};print months_diff}'

#!/usr/bin/env bash

# @(#) s1       Demonstrate a month duration between 2 dates, ddiff.

program=ddiff
program=dateutils.ddiff

# Utility functions: print-as-echo, print-line-with-visual-space, debug.
# export PATH="/usr/local/bin:/usr/bin:/bin"
LC_ALL=C ; LANG=C ; export LC_ALL LANG
pe() { for _i;do printf "%s" "$_i";done; printf "\n"; }
pl() { pe;pe "-----" ;pe "$*"; }
db() { ( printf " db, ";for _i;do printf "%s" "$_i";done;printf "\n" ) >&2 ; }
db() { : ; }
C=$HOME/bin/context && [ -f $C ] && $C $program

date1=03-02-2016
date2=15-11-2015

pl " Result difference in months:"
$program -i"%d-%m-%Y" "$date2" "$date1" -f "%m %d"

pl " A method for rounding up:"
read months days  < <( $program -i"%d-%m-%Y" "$date2" "$date1" -f "%m %d" )
if [ "$days" -gt 0 ]
then
  (( months++ ))
fi
pe " Rounded-up months is $months"

exit 0

$ ./s1

Environment: LC_ALL = C, LANG = C
(Versions displayed with local utility "version")
OS, ker|rel, machine: Linux, 3.16.0-4-amd64, x86_64
Distribution        : Debian 8.3 (jessie) 
bash GNU bash 4.3.30
dateutils.ddiff ddiff 0.3.1

-----
 Result difference in months:
2 19

-----
 A method for rounding up:
 Rounded-up months is 3