mjhjmtu

For a set of portable tools try my very own dateutils. Your two examples would boil down to one-liners:

ddiff 2011-11-15 2012-04-11
=>
 148

or in weeks and days:

ddiff 2011-11-15 2012-04-11 -f '%w %d'
=>
 21 1

and

dadd 2011-11-15 21w
=>
 2012-04-10

A python example for calculating the number of days I've walked the planet:

$ python
>>> from datetime import date as D
>>> print (D.today() - D(1980, 6, 14)).days
11476

I usually prefer having the time/date in unix utime format (number of seconds since the epoch, when the seventies begun, UTC). That way it always boils down to plain subtraction or addition of seconds.

The problem the usually becomes transforming a date/time into this format.

If you have GNU date, you can get it with date '+%s'
At the time of writing, the current time is 1321358027.

To compare with 2011-11-04 (my birthday), date '+%s' -d 2011-11-04, yielding 1320361200. Subtract: expr 1321358027 - 1320361200 gives 996827 seconds, which is expr 996827 / 86400 = 11 days ago.

The problem is converting from utime (1320361200 format) into a date. I don't know of a readily available tool to do this, but it's very simple to do in for instance C or perl.

This came up when using date -d "$death_date - $y years - $m months - $d days" to get a birth date (for genealogy). That command is WRONG. Months aren't all the same length, so (date + offset) - offset != date. Ages, in year/month/day, are measures going forwards from the date of birth.

$ date --utc -d 'mar 28 1867 +72years +11months +2days'
Fri Mar 1 00:00:00 UTC 1940

$ date --utc -d 'mar 1 1940 -72years -11months -2days'
Sat Mar 30 00:00:00 UTC 1867
# (2 days later than our starting point)

Date gives the correct output in both cases, but in the second case you were asking the wrong question. It matters WHICH 11 months of the year the +/- 11 cover, before adding/subtracting days. For example:

$ date --utc -d 'mar 31 1939 -1month'
Fri Mar 3 00:00:00 UTC 1939
$ date --utc -d 'mar 31 1940 -1month' # leap year
Sat Mar 2 00:00:00 UTC 1940
$ date --utc -d 'jan 31 1940 +1month' # leap year
Sat Mar 2 00:00:00 UTC 1940

For subtracting to be the inverse operation of adding, the order of operations would have to be reversed. Adding adds years, THEN months, THEN days. If subtracting used the opposite order, then you'd get back to your starting point. It doesn't, so you don't, if the days offset crosses a month boundary in a different length month.

If you need to work backwards from an end date and age, you could do it with multiple invocations of date. First subtract the days, then the months, then the years. (I don't think it's safe to combine the years and months in a single date invocation, because of leap years altering the length of February.)

If a graphical tool is OK for you, I heartily recommend qalculate (a calculator with an emphasis on unit conversions, it comes with a GTK and KDE interface, IIRC). There you can say e.g.

days(1900-05-21, 1900-01-01)

to get the number of days (140, since 1900 was not a leap year) between the dates, but of course you can also do the same for times:

17:12:45 − 08:45:12

yields 8.4591667 hours or, if you set the output to time formatting, 8:27:33.

I frequently use SQL for date calculations. For example MySQL, PostgreSQL or SQLite:

bash-4.2$ mysql <<< "select datediff(current_date,'1980-06-14')"
datediff(current_date,'1980-06-14')
11477

bash-4.2$ psql <<< "select current_date-'1980-06-14'"
 ?column? 
----------
 11477
(1 row)

bash-4.2$ sqlite2 <<< "select julianday('now')-julianday('1980-06-14');"
11477.3524537035

Other times I just feel in mood for JavaScript. For example SpiderMonkey, WebKit, Seed or Node.js:

bash-4.2$ js -e 'print((new Date()-new Date(1980,5,14))/1000/60/60/24)'
11477.477526192131

bash-4.2$ jsc-1 -e 'print((new Date()-new Date(1980,5,14))/1000/60/60/24)'
11477.47757960648

bash-4.2$ seed -e '(new Date()-new Date(1980,5,14))/1000/60/60/24'
11477.4776318287

bash-4.2$ node -pe '(new Date()-new Date(1980,5,14))/1000/60/60/24'
11624.520061481482

(Watch out when passing the month to the JavaScript Date object's constructor. Starts with 0.)

Another way to calculate the difference between two dates of the same calendar year you could use this:

date_difference.sh
1 #!/bin/bash
2 DATEfirstnum=`date -d "2014/5/14" +"%j"`
3 DATElastnum=`date -d "12/31/14" +"%j"`
4 DAYSdif=$(($DATElastnum - $DATEfirstnum))
5 echo "$DAYSdif"

• Line 1 declares to the shell which interpreter to use.


• Line 2 assigns the value from the out of date to the variable
  DATEfirstnum. The -d flag displays the string in a time format in
  this case May 14th 2014 and +"%j" tells date to format the output
  to just the day of the year (1-365).


• Line 3 is the same as Line 2 but with a different date and
  different format for the string, December 31st, 2014.


• Line 4 assigns the value DAYSdif to the difference of the two
  days.


• Line 5 displays the value of DAYSdif.

This works with the GNU version of date, but not on the PC-BSD/FreeBSD version. I installed coreutils from ports tree and used the command /usr/local/bin/gdate instead.

With the help of dannas solutions this can be done in one line with following code:

python -c "from datetime import date as d; print(d.today() - d(2016, 7, 26))"

(Works in both Python 2.x and Python 3.)

There's also GNU unit's time calculations combined with GNU date:

$ gunits $(gdate +%s)sec-$(gdate +%s -d -1234day)sec 'yr;mo;d;hr;min;s'
 3 yr + 4 mo + 16 d + 12 hr + 37 min + 26.751072 s
$ gunits $(gdate +%s -d '2015-1-2 3:45:00')sec-$(gdate +%s -d '2013-5-6 7:43:21')sec 'yr;mo;d;hr;min;s'
 1 yr + 7 mo + 27 d + 13 hr + 49 min + 26.206759 s

(gunits is units in Linux, gdate is date)

datediff.sh on github:gist

#!/bin/bash
#Simplest calculator two dates difference. By default in days

# Usage:
# ./datediff.sh first_date second_date [-(s|m|h|d) | --(seconds|minutes|hours|days)]

first_date=$(date -d "$1" "+%s")
second_date=$(date -d "$2" "+%s")

case "$3" in
"--seconds" | "-s") period=1;;
"--minutes" | "-m") period=60;;
"--hours" | "-h") period=$((60*60));;
"--days" | "-d" | "") period=$((60*60*24));;
esac

datediff=$(( ($first_date - $second_date)/($period) ))
echo $datediff

date and bash can do date differences (OS X options shown). Place the latter date first.

echo $((($(date -jf%D "04/03/16" +%s) - $(date -jf%D "03/02/16" +%s)) / 86400))
# 31

You can use the awk Velour library:

velour -n 'print t_secday(t_utc("2017-4-12") - t_utc("2017-4-5"))'

Or:

velour -n 'print t_secday(t_utc(ARGV[1]) - t_utc(ARGV[2]))' 2017-4-12 2017-4-5

Result:

7

Can someone explain those strange results ?????

$ date -d "10/30/2018 8:00:00 + 3 hours"
Tue Oct 30 07:00:00 CET 2018
$ date -d "10/30/2018 8:00:00 + 0 hours + 2 hours"
Tue Oct 30 12:00:00 CET 2018
$ date -d "10/30/2018 8:00:00 - 01:00"
Tue Oct 30 10:00:00 CET 2018