Date difference based on yyymm in unix - shell

Question

#####DATE1=201609
#### DATE2=201508
How to calculate the difference between these two date and get output as count of no of month
ie
201609-201508=13month

Answer 1

The calculation of the time difference is generally a complicated task, even for a single calendar type (and there are many). Many programming languages have a built-in support for date and time manipulation operations, including calculation of the time difference. But the most useful feature available in the popular shells is the date command which lacks this feature, unfortunately.
Therefore, we should whether write a script in another language, or make some assumptions such as the number of days in the year.
For example, in Perl the task is done with just four lines of code:
perl -e $(cat <<'PerlScript'
use Time::Piece;
my $t1 = Time::Piece->strptime($ARGV[0], '%Y%m');
my $t2 = Time::Piece->strptime($ARGV[1], '%Y%m');
printf "%d months\n", ($t1 - $t2)->months;
PerlScript
) 201609 201508
However, the difference of Time::Piece objects is an instance of Time::Seconds which actually assumes that
there are 24 hours in a day, 7 days in a week, 365.24225 days in a
year and 12 months in a year.
which indirectly confirms my words regarding the complexity of the task.
Then let's make the same assumption, and write a simple shell script:
DATE1=201609
DATE2=201508
printf '(%d - %d) / 2629744.2\n' \
$(date -d ${DATE1}01 +%s) \
$(date -d ${DATE2}01 +%s) | bc
where 2629744.2 is the number of seconds in month, i.e. 3600 * 24 * (365.24225 / 12).
Note, most of the shells do not support floating point arithmetic. That's why we need to invoke external tools such as bc.
The script outputs 13. This is a portable version. You may run it in the standard shell, Bash, Korn shell, or Zsh, for instance. If you want to put the result into a variable, just wrap the printf command in $( ... ):
months=$(printf '(%d - %d) / 2629744.2\n' \
$(date -d ${DATE1}01 +%s) \
$(date -d ${DATE2}01 +%s) | bc)
printf '%d - %d = %d months\n' $DATE1 $DATE2 $months

Answer 2

You can try below solution -
[vipin#hadoop ~]$ cat time.awk
{
diff1=((substr($1,1,4)) - (substr($2,1,4)))
diff2=((substr($1,5,2)) - (substr($2,5,2)))
if(diff1 > 0)
{
if(diff2 > 0)
{
print (diff1+diff2)
}
else if (diff2 = 0)
{
print (diff1+diff2)
}
else
{
diff2=(12+((substr($1,5,2)-(substr($2,5,2)))))
diff1=(diff1-1)
print (diff1+diff2)
}
}
else if(diff1 == 0)
{
if(diff2 > 0)
{
print (diff1+diff2)
}
else if (diff2 == 0)
{
print (diff1+diff2)
}
else
{
print "Argument 2 is greater than 1"
}
}
else
{
print "Argument 2 is greater than 1"
}
}
Test Cases -
[vipin#hadoop ~]$ cat time1.txt && awk -f time.awk time1.txt
201611 201601
10
[vipin#hadoop ~]$ cat time2.txt && awk -f time.awk time2.txt
201601 201611
Argument 2 is greater than 1
[vipin#hadoop ~]$ cat time3.txt && awk -f time.awk time3.txt
201511 201601
Argument 2 is greater than 1
[vipin#hadoop ~]$ cat time4.txt && awk -f time.awk time4.txt
201611 201611
0

Answer 3

Let's try something quick and dirty that won't work for all months in history, but it's probably good enough: convert YYYYMM to the number of months since year 0:
$ ym2m() {
if [[ $1 =~ ^([0-9]{4})([0-9]{2})$ ]]; then
echo $(( 10#${BASH_REMATCH[1]} * 12 + 10#${BASH_REMATCH[2]} ))
else
return 1
fi
}
$ ym2m 201609
24201
$ ym2m 201508
24188
$ echo $(( $(ym2m 201609) - $(ym2m 201508) ))
13
Notes:
requires bash version 4.3 (I think)
ym2m => "year-month to months"
uses 10#number in the arithmetic expression to ensure "08" and "09" are not treated as invalid octal numbers.