calculation of years, months and days

Hello

I have to calculate the duration in years, months and days between contract_start_date and contract_end_date

for example: Contract_Start_Date = 17 November 2006 "and Contract_End_Date = January 21, 2008"

Hello

have you tried the function months_between? You get the number of months between the dates, years and months is easy.
If you just truncates the result and we add_months with it and the date of departure, you can get a date from which you can get everyday simply by calculating the difference of the end date.

Then
years: = trunc (months_between (end_date, start_date) / 12);
month: = mod (months_between (start_date, end_date), 12);
days: = trunc (end_date - add_months (start_date, trunc (months_between (end_date, start_date)));)

It should work

Best regards
Wolfgang

Tags: Database

Similar Questions

How to get the difference in Date in years, months and days

Select Sysdate-Birth_date from Employees;

the query returns the number of days between two dates
I want to convert this number in years, months and days
Can someone help me please

Hoek wrote:
Try:

SQL> with t as (
2  select to_date('01-01-1980', 'dd-mm-yyyy') birthdate
3  ,      to_date('05-05-2010', 'dd-mm-yyyy') today
4  from   dual
5  )
6  --
7  --
8  --
9  select extract(year from today)-extract(year from birthdate) years
10  ,      extract(month from today)-extract(month from birthdate) months
11  ,      extract(day from today)-extract(day from birthdate) days
12  from   t;

YEARS     MONTHS       DAYS
---------- ---------- ----------
30          4          4

1 row selected.

(May need some adjustments, possibly using the LARGEST, but currently not as long)

Indeed, a few adjustments...

SQL> ed
Wrote file afiedt.buf

  1  with t as (
  2    select to_date('31-01-2009', 'dd-mm-yyyy') birthdate
  3    ,      to_date('28-02-2010', 'dd-mm-yyyy') today
  4    from   dual
  5    )
  6  --
  7  --
  8  --
  9  select extract(year from today)-extract(year from birthdate) years
 10  ,      extract(month from today)-extract(month from birthdate) months
 11  ,      extract(day from today)-extract(day from birthdate) days
 12* from   t
SQL> /

     YEARS     MONTHS       DAYS
---------- ---------- ----------
         1          1         -3

SQL>

Maybe something like this?

SQL> ed
Wrote file afiedt.buf

  1  with t as (select to_date('17-nov-2006','dd-mon-yyyy') as c_start_date, to_date('21-jan-2008','dd-mon-yyyy') as c_end_date from dual union all
  2             select to_date('21-nov-2006','dd-mon-yyyy'), to_date('17-feb-2008','dd-mon-yyyy') from dual union all
  3             select to_date('31-jan-2009','dd-mon-yyyy'), to_date('28-feb-2010','dd-mon-yyyy') from dual union all
  4             select to_date('21-jun-2006','dd-mon-yyyy'), to_date('17-jul-2008','dd-mon-yyyy') from dual
  5             )
  6  -- end of test data
  7  select c_start_date, c_end_date
  8        ,trunc(months_between(c_end_date, c_start_date) / 12) as yrs
  9        ,trunc(mod(months_between(c_end_date, c_start_date), 12)) as mnths
 10        ,trunc(c_end_date - add_months(c_start_date, trunc(months_between(c_end_date, c_start_date)))) as dys
 11* from t
SQL> /

C_START_DATE        C_END_DATE                 YRS      MNTHS        DYS
------------------- ------------------- ---------- ---------- ----------
17/11/2006 00:00:00 21/01/2008 00:00:00          1          2          4
21/11/2006 00:00:00 17/02/2008 00:00:00          1          2         27
31/01/2009 00:00:00 28/02/2010 00:00:00          1          1          0
21/06/2006 00:00:00 17/07/2008 00:00:00          2          0         26

How to find the difference in date in years, months and days
Hello
I need to find the difference between 2 dates in the following way "years months days". "
Please can help me, how can I achieve this.

for example, in the scott schema emp table, I need to find the difference in date between sysdate and hiredate for an employee in the following way.

12 years 7 months 4 days.
Hello

Please, see this post to AskTom [difference between 2 dates | http://asktom.oracle.com/pls/asktom/f?p=100:11:0:P11_QUESTION_ID:96012348060]. There is good information in this forum, for example you can also see this thread [calculation of years, months & days | http://forums.oracle.com/forums/thread.jspa?messageID=3115216�]

Kind regards

Published by: Walter Fernández on November 30, 2008 08:58 - adding another link

How to convert days to years, months and days remaining
Hi all
I have the number of days for example: 398 days how
convert 398 days to several years, and the number of months and days remaining
398 days 1 year, 1 month and 2 days
Concerning
Jocelyne
There is no correct answer.

That's right! ;)

Calendar.setTime defines the only year, month and day, no time
Hello

It is more a question of JAva instead of JDeveloper, but I still hope to get some answers here. I have problems, I have a Date (oracle.jbo.domain.Date) 2008-12-12 15:21:0. What I need is a string '121521' (day, hour, and minute - DDHHMM). This should be simple, but I'm strugling with this for a while now: $. I use class schedule so I can get the values using cal.get (Calendar.DAY_OF_MONTH), cal.get (Calendar.HOUR_OF_DAY) and cal.get (Calendar.MINUTE).

Think I tried and none of them work:
```
      Calendar cal = Calendar.getInstance();
          Date ldETA = (Date)loETA; // this is the oracle.jbo.domain.Date 12/12/2008 15:21:0

          cal.clear();
          cal.setTime(ldETA.dateValue());
         // I try getting the values here
          try {
            cal.clear();
          cal.setTimeInMillis(ldETA.longValue());
         // I try getting the values here
          }
          catch (Exception e) {}
          
          java.util.Date loD = (java.util.Date)ldETA.dateValue();
          cal.clear();
          cal.setTime(loD);
         // I try getting the values here
```
I always get only 12 for cal.get (Calendar.DAY_OF_MONTH) while cal.get (Calendar.MINUTE) and cal.get (Calendar.HOUR_OF_DAY) return 0.

Help please: (.)

Thank you
BB
BB,

oracle.jbo.domain.Date is a subclass of oracle.sql.DATE, which has a method toText() - have you tried?

John

Month and day of switching

I have a form that has fields Date1 and day 1 on it. When you enter the date Date1 must copy the date to the day 1 and change the format to display the day of the week. I tried this multiple ways and were not able to work right.
First, I tried the following:
Field: day 1
Format: Mm/dd/yyyy-ddd Date\Custom
Calculate: Custom calculation Script: event.value = this.getField("Date1").value
Result: The month and day are reversed... ex: 05/08/14 I entered Date1 and day1 shows Thursday, may 8, 2014
Then I tried:
Field day 1
Format: no
Calculate: Custom calculation Script: event.value = this.getField("Date1").value
Post: Run the custom validation script:
var f = event.value;
= event.value util.printd ("mmm dd/mm/yyyy", f);
Result: The date shows in Day1 but the format does not change
Finally, I tried:
Field day 1
Format: no
Calculate: Custom calculation Script:
var f = getField("Date1").value;
= event.value util.printd ("mmm dd/mm/yyyy", f);
Result: Nothing shows in day 1
Product: Adobe Acrobat Pro XI
OS: Windows 7 Professional
If anyone can help me I would be very happy
Thank you

You know anything about dates in JavaScript or any other computer language?

Dates and times for screens are text strings. To do many things with these channels, it must convert the object date JS. Then, there are a lot of properties and methods to deal with the date object. The conversion method also performs validation of date routines to ensure that the dates are the dates on the calendar.

Have you read the Acrobat JS Reference for the method of "util.printd"?

The util.printd has 3 parameters, the date format, the date object, and the logic to indicate the use of the image of XFA.

You only need the date and this date format string. The date object is not the chain of value of a date field.

You must use the 'util.scand' method to convert the string to a date in a date object.

Try the following script for 1 day field. The format should be 'None '.

get the date string;

var cDate1 = this.getField("Date_1").value;

date string format;

var cFormatDate1 = "mm/dd/yyyy";

convert the date string in the format of a date object.

var oDate1 = util.scand (cFormatDate1, cDate1);

error checking;

if(oDate1 == null) {}

App.Alert ("Error with date string" cDate1 "using the format" + cFormatDate1, 1, 0);

clear the value of the field;

Event.Value = "" ;// convert to date object;"

} else {}

format of the object date to display the day of the week;

Event.Value = util.printd ("ddd", oDate1);

}

To find the months and days between 2 dates

Hello
I want to find the months and days between 2 dates.
For example.
1 - Date: August 25, 2013
2 - Date: October 23, 2013
If we consider each month 30 days, it should give
August 25, 2013 to August 30, 2013 = 6 days
01-Sep-2013-30-Sep-2013 = 1 month
October 23, 2013 to October 30, 2013 = 8 days
Total = 1 month and 14 days.
Kindly help as soon as possible.
Thanks and greetings
Suresh

Assuming that d2 > d1,.

where d)

Select sysdate d1, sysdate + 56 double d2

Union all select to_date (March 1, 2013 ',' dd-mon-yyyy "") d1, to_date (March 31, 2013 ',' dd-mon-yyyy ') d2 double

Union all select to_date (5 February 2013 ',' dd-mon-yyyy ') d1, to_date (March 31, 2013 ',' dd-mon-yyyy "") double d2

Union all select to_date (February 25, 2013 ',' dd-mon-yyyy "") d1, to_date (March 23, 2013 ',' dd-mon-yyyy ') d2 double

Union all select to_date (February 25, 2013 ',' dd-mon-yyyy ') d1, to_date (March 31, 2013 ',' dd-mon-yyyy "") double d2

Union all select to_date (August 2, 2013 ',' dd-mon-yyyy "") d1, to_date (29 October 2013 ',' dd-mon-yyyy ') d2 double

Union all select to_date (February 1, 2013 ',' dd-mon-yyyy "") d1, to_date (May 31, 2013 ',' dd-mon-yyyy ') d2 double

Union all select to_date (25 August 2013 ',' dd-mon-yyyy "") d1, to_date ('03-Sep-2013', 'Mon-dd-yyyy') d2 double

Union all select to_date (July 30, 2013 ',' dd-mon-yyyy "") d1, to_date (August 31, 2013 ',' dd-mon-yyyy ') d2 double

Union all select to_date (July 31, 2013 ',' dd-mon-yyyy ') d1, to_date (August 30, 2013 ',' dd-mon-yyyy "") double d2

Union all select to_date (July 31, 2013 ',' dd-mon-yyyy ') d1, to_date (3 August 2013 ',' dd-mon-yyyy "") double d2

Union all select to_date (3 July 2013 ',' dd-mon-yyyy "") d1, to_date (August 31, 2013 ',' dd-mon-yyyy ') d2 double

Union all select to_date ('31-08-2013', ' dd-mm-yyyy'), to_date('05-10-2013','dd-mm-yyyy') of the double

Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('31-03-2013','dd-mm-yyyy') of the double

Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('05-03-2013','dd-mm-yyyy') of the double

Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('05-02-2013','dd-mm-yyyy') of the double

)

Select d1, d2,

1 + 30 * trunc (months_between (d2, d1)) + LESS (extract (day of d2), 30)-LESS (excerpt (d1 day), 30)

+ CASE when extracted (d2 day)< extract(day="" from="" d1)="" then="" 30="" else="" 0="" end ="">

d

D1 D2 DAYSBETWEEN

----------- ----------- -----------

October 10, 2013 5 December 2013 56

March 1, 2013 30 March 31, 2013

5 February 2013 March 31, 2013 56

February 25, 2013 March 23, 2013 29

February 25, 2013 March 31, 2013 36

August 2, 2013 29 October 2013 88

February 1, 2013 may 31, 2013 120

August 25, 2013 03 - Sep-2013 9

July 30, 2013 31 August 31, 2013

July 31, 2013 August 30, 2013 31

July 31, 2013 3 August 2013 4

July 3, 2013 August 31, 2013 58

31 August 2013 5 October 2013 36

5 February 2013 March 31, 2013 56

5 February 2013 March 5, 2013 31

February 5, 2013 February 5, 2013 1

In my view, which corresponds to your rules.

Months and days
Hello world

I have here a difficult problem.

I have documents that contain data for several months. For example...

inv_no datefrom, dateto amount corresponding
1234 21 January 09 xxx 18 March 09

... and, in order to reconcile the billing, I need to split the amount and assign it to the appropriate depending on the day billed during this particular month to month. i.e.

corresponding inv_no component datefrom, dateto NoDays amount
1234 a January 21 09 31 January 09 11 xxx
1234 B 1 February 09 28 February 09 28 xxx
1234 C 1st March 09 18 March 09 18 xxx

I did experiment with add_months and subtracting one date from another, but with different numbers of days in particular months, it gets very complicated.

So I thought maybe it's a problem that appears quite often, and perhaps Oracle (SQL or PL/SQL?) has specific functions to handle this kind of thing.

I don't want to reinvent the wheel :-)

Advice would be a great help.

Best regards and thanks,
Alan Searle
I don't know if there is something built into Oracle.

Is that what you are looking for?
```
SQL> with t as
  2  (
  3  select '1234' inv_no, to_date('21-jan-2009', 'DD-MON-YYYY') date_from, to_date('18-mar-2009', 'DD-MON-YYYY') date_to, 1000 amount from dual
  4  )
  5  select inv_no
  6         ,decode(level, 1, date_from, trunc(add_months(date_from, level -1), 'MM')) date_from
  7         ,last_day(decode(level, 1, date_from, trunc(add_months(date_from, level -1), 'MM'))) date_to
  8         ,amount*(last_day(decode(level, 1, date_from, trunc(add_months(date_from, level -1), 'MM'))) - decode(level, 1, date_from, trunc(add_months(date_from, level -1), 'MM')))/(date_to - date_from) amt
  9  from t
 10  connect by level <= to_char(date_to, 'MM') - to_char(date_from, 'MM') +1
 11  /

INV_ DATE_FROM DATE_TO        AMT
---- --------- --------- --------
1234 21-JAN-09 31-JAN-09      179
1234 01-FEB-09 28-FEB-09      482
1234 01-MAR-09 31-MAR-09      536

SQL>
```
See you soon
Sarma.

split with day, month and year in a date

Hi all
I have this point of view
with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
)
--
-end of test data
--
SELECT id, dt
floor (months_between(sysdate,dt)/12) years
, floor (mod (months_between (sysdate, dt), 12)) months
-The day is ambiguous because of the number of days in a different month
-you could design your own algorithm to give the desired results, but it is a good approximation
, floor (sysdate-(add_months (dt, floor (months_between (sysdate, dt))) as dys
t
/
ID DT YEARS MONTHS DYS
---------- ----------- ---------- ---------- ----------
1-6 FEBRUARY 2010 5 8 22
2 29 NOVEMBER 2001 13 10 29
3-6 FEBRUARY 2011 4 8 22
4-10 OCTOBER 2011 4 0 18
As a result, 26 years old, 26 months, 91 days.

And subtract years, MONTHS and DYS as format
erase years, from 1 to 12 months and the days of the 1 to 30 as
91 days = 30 x 3 + 1 = > 1 day over 3 months
26 months + 3 months = 29 months = > 24 months + 5 months = 2 years + 5 months
and 26 years + 2 years = 28
desire to result: 28 years, 5 months and 1 day.
Is there a function that give me this result from the query above?
Kind regards
Gordan

Hello Gordan,.

I understand that you are the SUM of all the lines.

Here are two ideas:

-1 - using a date
(a) SUM (TRUNC (sysdate) - dt)
I would give you the total number of days.
(b) adding the number of days to, for example, DATE ' 1900-01-01' would give another date...
(c) simply subtract the years 1900 and it takes several years,
subtract 1 from the month (1-12 from January to December) and you have the number of months
subtract 1 from the day of the month (1-31) and you have the number of days.
It is of course 'approximate' as the months do not have the same number of days...

-2 - using only the number of days
Maybe you want something else, for example: the total number of days divided by 365.25 for many years, the recall divided by 30 to get the number of months, the reminder being the number of days.

(a) SUM (TRUNC (sysdate) - dt)

I would give you the total number of days.

(b) Division by 365.25 (le.25 is to take leap years into account, but you can choose for example "365")

(c) number of reminder of days (total - years * 365.25): divided by 30 is the number of months

(d) number reminder of days (total - years * 365.25 - months * 30): gives the number of days.

The following two options with your test data (by chance, this gives the same result in this case)

option 1:
with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
)
target_date AS
(SELECT DATE ' 1900-01-01' + SUM (TRUNC (sysdate) - t.dt) FROM t trgt)
in_pieces AS
(SELECT TO_NUMBER (TO_CHAR (td.trgt, 'YYYY')) - 1900 y
, TO_NUMBER (TO_CHAR (td.trgt, 'MM')) - 1 m
, TO_NUMBER (TO_CHAR (td.trgt, 'DD')) - 1 d
Target_date TD
)
SELECT "result: ' |"
TRIM (CASE WHEN ip.y = 0 THEN NULL
WHEN ip.y = 1 THEN "1 year"
Of OTHER TO_CHAR (ip.y) | "years".
END |
CASE WHEN ip.m = 0 THEN NULL
WHEN ip.m = 1 THEN "1 month"
Of OTHER TO_CHAR (ip.m) | 'months '.
END |
CASE WHEN ip.d = 0 THEN NULL
WHEN ip.d = 1 THEN "1 day"
Of OTHER TO_CHAR (ip.d) | 'days '.
END
)
Of in_pieces ip
;
result: 28 years, 4 months and 28 days

option 2:

with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
)
nb_days AS
(SELECT SUM (TRUNC (sysdate) - t.dt) n t)
, y AS
(SELECT Nb.n, FLOOR (nb.n / 365.25) y nb_days n. b.)
ym AS
(SELECT y.n, y.y, FLOOR ((y.n-y.y * 365,25) / 30) FROM a)
ymd AS
(SELECT ym.y, ym.m, ym.n - ym.y * 365.25 - ym.m * ym FROM 30 d)
SELECT "result: ' |"
TRIM (CASE WHEN ymd.y = 0 THEN NULL
WHEN ymd.y = 1 THEN "1 year"
Of OTHER TO_CHAR (ymd.y) | "years".
END |
CASE WHEN ymd.m = 0 THEN NULL
WHEN ymd.m = 1 THEN "1 month"
Of OTHER TO_CHAR (ymd.m) | 'months '.
END |
CASE WHEN ymd.d = 0 THEN NULL
WHEN ymd.d = 1 THEN "1 day"
Of OTHER TO_CHAR (ymd.d) | 'days '.
END
)
WAN
;
result: 28 years, 4 months and 28 days

Best regards

Bruno Vroman.

How to find the next date of the year and day as of today's date and the day

I have a question about date functions, that is to say: how to get one next year (date and day) like today are the date and the day this year?

You mean like this?

SQL > select to_char (add_months (sysdate, 12),' day DD/MM/YYYY ') twice;

TO_CHAR (ADD_MONTHS (S
--------------------
Friday, August 22, 2014

Get the number of days in a month based on the month and year of fields

I have a column in my form which lists the days in a month. I want to configure a hidden field that calculates the total number of days in a month, based on the month and year of the field inputs. The number of days will determine what appears on the column. For example, if I put 4 months, and 2016 in the field of the year, I get 30 in the hidden field. Thus, on the column 'Day', I'll have numbers 1-30. Or if I put 2 months and 2016 in the field of the year, I get the 29 in the hidden field. If the numbers 1-29 appears in the column 'day '.
Found this on some forum javascript code:
//Month is 1 based function daysInMonth(month,year) { return new Date(year, month, 0).getDate(); } //July daysInMonth(7,2009); //31 //February daysInMonth(2,2009); //28 daysInMonth(2,2008); //29
I do not know how to convert this code in JavaScript to adobe and don't really know how to use it. All I know how to do is to configure the field values for the field month and year as variables. I am a novice programmer and would appreciate it really all the help I can get. Thank you in advance!

The code seems to be JavaScript and runs as needed by using the JavaScript console.

I would like to consider making more general code, so if you have a date string that includes at least the month and year we could just call the function and get the number of days for that month.

The following script will calculate the number of days in a month, by using at least the month and year values can display the result on the JavaScript console and all of the value field for the field that has this code as the custom calculation Script.

function daysInMonth (oDate) {}
return new Date (oDate.getFullYear (), oDate.getMonth () + 1, 0) .getDate ();
}

nMonth var = this.getField("Month").valueAsString; get the value of month;
nYear var = this.getField("Year").valueAsString; get the value of the year;

Event.Value = "";

If (nMonth! = "" & nYear!) = "") {}
var MyDate = util.scand ("' / mm/yyyy ', nMonth +" / "+ nYear); convert to date object;
var nDaysInMonth = daysInMonth (MyDate); get the number of days;

Console.Open (); Open the JavaScript console;

Console.clear(); clear the console;

Console.println ("Days in" + nMonth + ":" + nDaysInMonth); show days in month;

Event.Value = nDaysInMonth; Set the value of the field;

}

Get the number of days, months and year
Hi all

I'm using OBIEE 11 g. I have a line of dash for year (2010, 2011, etc.) and months (Jan, Feb,..., Dec). How can I get the number of days in a given month and year in my report?

For example, if I choose to Jan & 2011, I'd get 31 and 356. And if I choose Feb & 2011, I would get 28 and 356.

Thank you!
I have something more good option here... Try to use this

DAYOFMONTH (TIMESTAMPADD (SQL_TSI_DAY, DAYOFMONTH (CURRENT_DATE) *-1, TIMESTAMPADD (SQL_TSI_MONTH, 1, CURRENT_DATE)))

This will give you the days for the month in progress (Oct 2011) after calculating the last day of each month... Similary dynamically pass the value Date or month... You will need to modify this function but shud be simple. Hope this helps :)

I want to collect rain for the day, month and year; Can what formula I use?

I want to collect rain for the day, month and year; What formula or expression that I can use.

I use a Rain Wise product that converts impulses to an analog value. The rain wise device can be

"the value measure up to 1", 5", or 10". I will be setting the unit at 10 inches in increments of 0.01 inch.

What I would do is at each time increments of signal I would consider it as 0.01 then after reaching

a period of time a day or a month, back to zero.

Need some advice on this problem.

How about you, using a counter to count the number of times where the signal is less than 0.01?

For example,.

Counter.Count = signal<=>

The total number of rain is

Counter * 10 + signal

Use TimeOfDay or TimeOfWeek to reset the counter.

Set the calendar to display day, month and year

In addition to the time of day, my computer began giving the day of the week (Monday) only instead of day, month and year as it did before. Tried to go back to 29/12/2014 or something similar without success. Any suggestions?

Hello

Sorry for the late reply and I appreciate your patience.

User profile corruption might have caused the problem. As a solution, I suggest to create a new user profile and check if the date format displays the same thing.

To do this, see the link below.

http://Windows.Microsoft.com/en-us/Windows/fix-corrupted-user-profile#1TC=Windows-7.

It will be useful. Just answer us with the State, and we will be happy to help you.

Pedometer in health app inaccurate since ios10.02 update and followed by the month and the year disappeared. Why?

I used the pedometer in the application of health that accompanies the iPhone6 for 18 months. I walk on a treadmill each morning for the same distance and time. The number of step was close enough for more than a year, but then I updated to ios10.02 and the count has been reduced by more than 2500 steps every day for the same duration of time and distance on the treadmill. He remained much lower now for two weeks. I also notice that the dashboard no longer offers a total annual and medium, or monthly. Is there a fix to come or was the unit off so much for the previous year?

kjm2222 wrote:

I also notice that the dashboard no longer offers a total annual and medium, or monthly.

You can still access graphics monthly and annual. In the form of health data, press activity. Swipe to today. Tap measures. You will then see a graph and a bar with options to choose to see it in days, months, or year. This works with any metric system.

calculation of years, months and days

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