mjhjmtu

Is there any simple way to summarize a day value like (1327 days) to the format:

xx years; xx month; xx days

without having to use a separate variable for each value.

Preferably with one command.

For a duration that includes a number of months or years, that has to make reference to a particular date, as different months or different years have different lengths.

To know how many years/months/days from now to 1327 days from now, with dateutils:

$ ddiff -f '%Y years, %m months, %d days' today "$(dadd now 1327)"
3 years, 7 months, 19 days

(you may sometimes find ddiff available as datediff or dateutils.ddiff; same for dadd).

That's what I get now on 2017-09-25 (because that's from 2017-09-25 to 2021-05-14). If I were to run that on 2018-03-01, I'd get:

3 years, 7 months, 17 days

because that's from 2018-03-01 to 2021-10-18.

And on that 2018-03-01 day, 1327 days ago would give 3 years, 7 months, 16 days.

More info at How can I calculate and format a date duration using GNU tools, not a result date?

I think that's a bit too complicated for a reasonably accurate solution (i. e. regarding calendar oddities like leap days) in Bash. Try something with a programming library for calendars like Python instead:

#!/usr/bin/env python3
import sys, calendar
from datetime import *

difference = timedelta(days=int(sys.argv[1]))
now = datetime.now(timezone.utc).astimezone()
then = now - difference

years = now.year - then.year
months = now.month - then.month
days = now.day - then.day
if days < 0:
 days += calendar.monthrange(then.year, then.month)[1]
 months -= 1
if months < 0:
 months += 12
 years -= 1

print(' year(s); month(s); day(s)'.format(years, months, days))

Example invocation:

$ ./human-redable-date-difference.py 1327
3 year(s); 7 month(s); 19 day(s)

Of course you can adjust the input and output format to your liking to select time differences based on other things than the number of days.