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Posted 25 Aug 2008


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Calculating Duration Between Two Dates in Years, Months and Days

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25 Aug 2008CPOL5 min read
Calculate the accurate duration between two dates in years, months and days


I was looking for a Duration calculator between two dates which gives duration in number of years, number of months and number of days together. It can be defined as Age Calculator which is able to give any one’s exact age in years, months and days.

But unfortunately I didn't find exactly what I was looking for. Most of the calculators give output either in years or in months or in weeks or in days and so on and even in milliseconds, but not all the information together.

I realize that there should be something to calculate duration between two dates and give output in years, months and days together.


In most of the solutions for this problem, I found that the difference is calculated by the following procedure:

First calculate ‘TimeSpan’ from difference between two dates. Then from the time span, calculate days, from days calculate months, and from months or days calculate years.

This calculates right up to day calculation. But when you calculate month from day, then there is conflict. Because, the length of all the months (length means days) is not the same. Some months contain 28 days, some 30 and 31 for regular years. So, two months could be equivalent to 59 (January + February) days, 61 (March + April) days, 62 (July + August) days and even 60 (February leap year + March) days for leap year. After that when calculating years, there is another conflict, does the year contain 365 days or 366 days (in case of leap year)?

Actually this problem can be easily solved for a single year by adding many conditions. But for a long duration, it is difficult to trace all the leap years and finally generate accurate duration in days, months and years.


Before starting the calculation, we just try to recollect our arithmetic operation basics. When we subtract 19 from 23 (23-19) then what happens? First we try to subtract the first digit ‘9’ (right side of 19) from first digit ‘3’ (right side of 23). Then we find that 3 is less than 9. So, add 10 with ‘3’ and subtract ‘9’ from (3+10); after that we add ‘1’ with ‘1’ second digit of 19 and subtract from the second digit ‘2’ of 23. We add 10 because the base of all the digits is 10.

In the same way, when we try to subtract one date from another (ignoring time) then we have to consider three different type/based numbers, that is day of the month; month of the year and year itself. Let's say we want to subtract date2 from date1 [date format YYY-mm-dd].

  • date1: 2000- 3- 1
  • date2: 1999- 4-10
  • dura: 0y - 10m - 21d

Here date1 is greater than date2, but day of the date1 ‘1’ is smaller then the day of the date2 ‘10’. So according to the arithmetic rule, before subtracting ‘10’ from ‘1’, we have to add a certain number (say x) with ‘1’ to the day of date1. Here x is equivalent to days (considering days is base of a month) of the month of date2. And also add ‘1’ with ‘4’ the month of date2. The number (x) will be different for different months, because different months contain different number of days. For example: for January x = 31, for February x= 28 or 29(for leap year) and so on.

After calculating days, we subtract the month of date2 ‘4’ including ‘1’ (according to day calculation) that is ‘4+1’ from ‘3’. As month of date1 (4+1=5) is less than the month of date2 ‘3’, we have to add 12 with month of date1 and add ‘1’ with year of date2. We are adding ‘12’ with month of date1 because each year contains exactly 12 months.

Finally, the year calculation is a simple arithmetic calculation.

Using the Code

Duration Calculation

Global Variables

private int[] monthDay = new int[12] { 31, -1, 31, 30, 31, 30, 31, 31, 30, 31, 30,
31 };
private DateTime fromDate;
private DateTime toDate;
private int year;
private int month;
private int day; 
  • int[] monthDay’ defines Number of days in month, index 0=> January and 11=> December. February contain either 28 or 29 days, that's why here value is -1 which means it will be calculated later.
  • DateTime fromDate’ contain the start date value or smaller date value between two dates.
  • DateTime toDate’ contains the end date value or bigger date value between two dates.
  • int year’ contains year(s) of the output.
  • int month’ contains month(s) of the output.
  • int day’ contains day(s) of the output.

Prepare Data for Calculation

To calculate duration, we need two dates:

public DateDifference(DateTime d1, DateTime d2)

Here d1 is the first date and d2 is the second date.

if (d1 >d2)
    this.fromDate = d2;
    this.toDate = d1;
    this.fromDate = d1;
    this.toDate = d2;

From the two dates, we identify which date is bigger. The bigger one is set as toDate and the smaller one is fromDate so that we always get a positive duration.


Day Calculation
increment = 0; 
if (this.fromDate.Day > this.toDate.Day)
    increment = this.monthDay[this.fromDate.Month - 1]; 

If ‘this.fromDate.Day’ is greater than ‘this.toDate.Day’, then we store the value in ‘increment’ which will be added to ‘this.toDate.Day’. To get the proper number ‘int[] monthDay’ will help us.

if (increment== -1)
    if (DateTime.IsLeapYear(this.fromDate.Year))
        increment = 29;
        increment = 28;

Here we check if the month is February? If it is, then what are the number of days?

if (increment != 0)
    day = (this.toDate.Day+ increment) - this.fromDate.Day;
    increment = 1; 
    day = this.toDate.Day - this.fromDate.Day;

The simple arithmetic operation is completed. And now ‘increment’ contains the number which will be added to ‘this.fromDate.Month’.

Up to this day, calculation is completed and ‘day’ contains the output’s day(s) result.

Month Calculation
if ((this.fromDate.Month + increment) > this.toDate.Month)
    this.month = (this.toDate.Month+ 12) - (this.fromDate.Month + increment);
       increment = 1;
    this.month = (this.toDate.Month) - (this.fromDate.Month + increment);
    increment = 0;

Month calculation is very simple and almost like Day calculation. Here if ‘this.toDate.Month’ is smaller than the result of ( ‘this.fromDate.Month’ + ‘increment’), then just add ‘12’ to ‘this.toDate.Month’ and add ‘1’ to ‘this.fromDate.Year’ (which is stored in ‘increment’); Otherwise, it is just simple subtraction.

Here we add 12 because each year contains exactly 12 months.

Year Calculation
this.year = this.toDate.Year - (this.fromDate.Year + increment);

This is just simple arithmetic operation. Nothing to say.

Final Results
public int year;
public int month;
public int day;

These are the variables which contain the results of the calculation.

public override string ToString()
    return this.year + "Year(s), " + this.month + " month(s), " + + " day(s)";

To get the formatted output, we override the tostring method.


Here is the output for the corresponding input. Input Date format is (yyyy-mm-dd) for user friendly view.

Output: 1 Year(s), 9 month(s), 26 day(s)  

Output: 24 Year(s), 6 month(s), 6 day(s)  

Output: 48 Year(s), 1 month(s), 0 day(s)  

Output: 48 Year(s), 1 month(s), 0 day(s)  

Output: 48 Year(s), 2 month(s), 8 day(s) 


This article will help you to find out the duration of dates in year, month and day format. The whole code is uploaded here, so anyone can play with it. And if there are any suggestions or corrections, please feel free to share.


  • 25th August, 2008: Initial post


This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)


About the Author

Mohammed Ali Babu
Software Developer AIUB
Bangladesh Bangladesh
No Biography provided

Comments and Discussions

QuestionExcelent! Thnks Pin
Prof. Francisco Araújo19-Feb-21 8:22
MemberProf. Francisco Araújo19-Feb-21 8:22 
QuestionIn Java Please Pin
Urek24-Jul-18 1:57
MemberUrek24-Jul-18 1:57 Pin
Member 1340912114-Sep-17 17:11
MemberMember 1340912114-Sep-17 17:11 
QuestionSame example in VBA Pin
Francisco Prado29-Apr-17 5:39
MemberFrancisco Prado29-Apr-17 5:39 
AnswerRe: Same example in VBA Pin
Ajos121-May-17 7:32
MemberAjos121-May-17 7:32 
AnswerI Need your Help ASAP Pin
Member 1340912114-Sep-17 17:54
MemberMember 1340912114-Sep-17 17:54 
PraiseBest answer I can get on the internet! Pin
Member 1053215116-Mar-17 23:27
MemberMember 1053215116-Mar-17 23:27 
QuestionIn VB please. Pin
Member 1240866424-Mar-16 10:42
MemberMember 1240866424-Mar-16 10:42 
Questionnot accurate calculations Pin
Patil Kishor10-Dec-13 23:59
MemberPatil Kishor10-Dec-13 23:59 
QuestionMore easy Pin
Gustavo Santis13-Mar-13 17:28
MemberGustavo Santis13-Mar-13 17:28 
AnswerRe: More easy Pin
Dinesh.V.Kumar29-Nov-13 2:17
MemberDinesh.V.Kumar29-Nov-13 2:17 
Questionty Pin
dipan chikani25-Aug-12 10:00
Memberdipan chikani25-Aug-12 10:00 
GeneralYou are great!! Pin
AviMor24-May-12 12:08
MemberAviMor24-May-12 12:08 
GeneralYou are great!! Pin
AviMor24-May-12 12:07
MemberAviMor24-May-12 12:07 
GeneralMy vote of 5 Pin
Mardani Dani5-Jan-12 13:08
MemberMardani Dani5-Jan-12 13:08 
GeneralMy Vote of 5 Pin
RaviRanjanKr1-Dec-11 5:08
professionalRaviRanjanKr1-Dec-11 5:08 
GeneralMy vote of 5 Pin
ben jamir30-Nov-11 21:59
Memberben jamir30-Nov-11 21:59 
QuestionThanks [modified] Pin
mahesh__kumar__sharma9-Jul-11 2:58
Membermahesh__kumar__sharma9-Jul-11 2:58 
Thanks this solved my problem. But it give not accureate result Frown | :(

i implement some different logic: it is as
A : Get the difference b/w both date.
B: Add difference in base date.
B 1) base date: year 0001, month : as from date's month: day as from date's day
B 2) if from date is leap year than base date: year 0004, month : as from date's month: day as from date's day
C: check if from_date's month is greater than to date than subtract 1 year from from_date & add 12 months.
D: check if from_date's day is greater than to date than subtract 1 month from from_date & add days of current month of to_date.

code is on this linkl[^]

-- Modified Sunday, July 17, 2011 1:37 PM

AnswerRe: Thanks Pin
acko0856-Sep-11 14:01
Memberacko0856-Sep-11 14:01 
AnswerRe: Thanks (issue with leap year) ... Pin
xirc_za12-Mar-13 13:57
Memberxirc_za12-Mar-13 13:57 
AnswerRe: Thanks [modified] Pin
MeenakshiLuthra22-Jan-12 23:37
MemberMeenakshiLuthra22-Jan-12 23:37 
GeneralSmall correction Pin
acko0857-May-11 19:29
Memberacko0857-May-11 19:29 
GeneralGood work - made few changes to make it more accurate Pin
Juvenile8-Dec-10 1:17
MemberJuvenile8-Dec-10 1:17 
GeneralMy vote of 5 Pin
masterJalchr25-Nov-10 22:10
MembermasterJalchr25-Nov-10 22:10 
GeneralHours, Mins, Seconds as well. Pin
Armoghan Asif23-Nov-10 6:26
MemberArmoghan Asif23-Nov-10 6:26 

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