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# Calculating Duration Between Two Dates in Years, Months and Days

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

## Introduction

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.

## Problem

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.

## Background

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;
}
else
{
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.

#### Calculation

##### 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;
}
else
{
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;
}
else
{
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;
}
else
{
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), " + this.day + " day(s)";
} ```

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

## Output

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

```Input:
2000-12-1
1999-2-3
Output: 1 Year(s), 9 month(s), 26 day(s)

Input:
1984-2-14
2008-8-20
Output: 24 Year(s), 6 month(s), 6 day(s)

Input:
2008-7-14
1960-6-14
Output: 48 Year(s), 1 month(s), 0 day(s)

Input:
2008-7-14
1960-6-14
Output: 48 Year(s), 1 month(s), 0 day(s)

Input:
2008-7-13
1960-5-5
Output: 48 Year(s), 2 month(s), 8 day(s) ```

## Conclusion

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.

## History

• 25th August, 2008: Initial post

## About the Author

 Software Developer AIUB Bangladesh
No Biography provided

## Comments and Discussions

 First PrevNext
 In Java Please Urek24-Jul-18 1:57 Urek 24-Jul-18 1:57
 VB.net Member 1340912114-Sep-17 17:11 Member 13409121 14-Sep-17 17:11
 Same example in VBA Francisco Prado29-Apr-17 5:39 Francisco Prado 29-Apr-17 5:39
 Re: Same example in VBA Ajos121-May-17 7:32 Ajos1 21-May-17 7:32
 I Need your Help ASAP Member 1340912114-Sep-17 17:54 Member 13409121 14-Sep-17 17:54
 Best answer I can get on the internet! Member 1053215116-Mar-17 23:27 Member 10532151 16-Mar-17 23:27
 In VB please. Member 1240866424-Mar-16 10:42 Member 12408664 24-Mar-16 10:42
 not accurate calculations Patil Kishor10-Dec-13 23:59 Patil Kishor 10-Dec-13 23:59
 More easy Gustavo Santis13-Mar-13 17:28 Gustavo Santis 13-Mar-13 17:28
 Days: endDate.Subtract(firstDate).TotalDays; or endDate.Subtract(firstDate).Days; Months: ((endDate.Year - firstDate.Year) * 12) + (endDate.Month - firstDate.Month); Years: endDate.Year - firstDate.Year;
 Re: More easy Dinesh.V.Kumar29-Nov-13 2:17 Dinesh.V.Kumar 29-Nov-13 2:17
 ty dipan chikani25-Aug-12 10:00 dipan chikani 25-Aug-12 10:00
 You are great!! AviMor24-May-12 12:08 AviMor 24-May-12 12:08
 You are great!! AviMor24-May-12 12:07 AviMor 24-May-12 12:07
 My vote of 5 Mardani Dani5-Jan-12 13:08 Mardani Dani 5-Jan-12 13:08
 My Vote of 5 RaviRanjanKr1-Dec-11 5:08 RaviRanjanKr 1-Dec-11 5:08
 My vote of 5 ben jamir30-Nov-11 21:59 ben jamir 30-Nov-11 21:59
 Thanks [modified] mahesh__kumar__sharma9-Jul-11 2:58 mahesh__kumar__sharma 9-Jul-11 2:58
 Re: Thanks acko0856-Sep-11 14:01 acko085 6-Sep-11 14:01
 Re: Thanks (issue with leap year) ... xirc_za12-Mar-13 13:57 xirc_za 12-Mar-13 13:57
 Re: Thanks [modified] MeenakshiLuthra22-Jan-12 23:37 MeenakshiLuthra 22-Jan-12 23:37
 Small correction acko0857-May-11 19:29 acko085 7-May-11 19:29
 Good work - made few changes to make it more accurate Juvenile8-Dec-10 1:17 Juvenile 8-Dec-10 1:17
 My vote of 5 masterJalchr25-Nov-10 22:10 masterJalchr 25-Nov-10 22:10
 Hours, Mins, Seconds as well. Armoghan Asif23-Nov-10 6:26 Armoghan Asif 23-Nov-10 6:26
 Excellent job samMaster1-Dec-09 21:09 samMaster 1-Dec-09 21:09
 Last Visit: 20-Feb-20 18:59     Last Update: 20-Feb-20 18:59 Refresh 12 Next »

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Article
Posted 25 Aug 2008

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