Difference in Months Between Two Dates

function monthDiff(d1, d2) {

    var months;

    months = (d2.getFullYear() - d1.getFullYear()) * 12;

    months -= d1.getMonth();

    months += d2.getMonth();

    return months <= 0 ? 0 : months;

}


function test(d1, d2) {

    var diff = monthDiff(d1, d2);

    console.log(

        d1.toISOString().substring(0, 10),

        "to",

        d2.toISOString().substring(0, 10),

        ":",

        diff

    );

}


test(

    new Date(2008, 10, 4), // November 4th, 2008

    new Date(2010, 2, 12)  // March 12th, 2010

);

// Result: 16


test(

    new Date(2010, 0, 1),  // January 1st, 2010

    new Date(2010, 2, 12)  // March 12th, 2010

);

// Result: 2


test(

    new Date(2010, 1, 1),  // February 1st, 2010

    new Date(2010, 2, 12)  // March 12th, 2010

);

// Result: 1

calculating the difference in months between two dates

You won't be able to get that from a TimeSpan, because a "month" is a variable unit of measure. You'll have to calculate it yourself, and you'll have to figure out how exactly you want it to work.

For example, should dates like July 5, 2009 and August 4, 2009 yield one month or zero months difference? If you say it should yield one, then what about July 31, 2009 and August 1, 2009? Is that a month? Is it simply the difference of the Month values for the dates, or is it more related to an actual span of time? The logic for determining all of these rules is non-trivial, so you'll have to determine your own and implement the appropriate algorithm.

If all you want is simply a difference in the months--completely disregarding the date values--then you can use this:

public static int MonthDifference(this DateTime lValue, DateTime rValue)
{
    return (lValue.Month - rValue.Month) + 12 * (lValue.Year - rValue.Year);
}

Note that this returns a relative difference, meaning that if rValue is greater than lValue, then the return value will be negative. If you want an absolute difference, you can use this:

public static int MonthDifference(this DateTime lValue, DateTime rValue)
{
    return Math.Abs((lValue.Month - rValue.Month) + 12 * (lValue.Year - rValue.Year));
}

Calculating number of full months between two dates in SQL

The original post had some bugs... so I re-wrote and packaged it as a UDF.

CREATE FUNCTION FullMonthsSeparation 
(
    @DateA DATETIME,
    @DateB DATETIME
)
RETURNS INT
AS
BEGIN
    DECLARE @Result INT

    DECLARE @DateX DATETIME
    DECLARE @DateY DATETIME

    IF(@DateA < @DateB)
    BEGIN
        SET @DateX = @DateA
        SET @DateY = @DateB
    END
    ELSE
    BEGIN
        SET @DateX = @DateB
        SET @DateY = @DateA
    END

    SET @Result = (
                    SELECT 
                    CASE 
                        WHEN DATEPART(DAY, @DateX) > DATEPART(DAY, @DateY)
                        THEN DATEDIFF(MONTH, @DateX, @DateY) - 1
                        ELSE DATEDIFF(MONTH, @DateX, @DateY)
                    END
                    )

    RETURN @Result
END
GO

SELECT dbo.FullMonthsSeparation('2009-04-16', '2009-05-15') as MonthSep -- =0
SELECT dbo.FullMonthsSeparation('2009-04-16', '2009-05-16') as MonthSep -- =1
SELECT dbo.FullMonthsSeparation('2009-04-16', '2009-06-16') as MonthSep -- =2

How to get the difference in months between two dates in PHP

$currentDateTime = new DateTime;
$dateTimeInTheFuture = new DateTime(MyObject->getDate());

$dateInterval = $dateTimeInTheFuture->diff($currentDateTime);

$totalMonths = 12 * $dateInterval->y + $dateInterval->m;

echo $totalMonths;

Calculate the number of months between two dates in PHP?

$date1 = '2000-01-25';
$date2 = '2010-02-20';

$ts1 = strtotime($date1);
$ts2 = strtotime($date2);

$year1 = date('Y', $ts1);
$year2 = date('Y', $ts2);

$month1 = date('m', $ts1);
$month2 = date('m', $ts2);

$diff = (($year2 - $year1) * 12) + ($month2 - $month1);

You may want to include the days somewhere too, depending on whether you mean whole months or not. Hope you get the idea though.

Pandas - Number of Months Between Two Dates

Here is a very simple answer my friend:

df['nb_months'] = ((df.date2 - df.date1)/np.timedelta64(1, 'M'))

and now:

df['nb_months'] = df['nb_months'].astype(int)

Calculate how many n.months between two dates

Thanks for @Cary's and @Lam's answer.

Here is my answer to find the n month.

# try to find the minimum n month between start_date and target_date
def calc_diff(start_date, target_date)
  months_diff = (target_date.year * 12 + target_date.month) - (start_date.year * 12 + start_date.month)

  ## need to check the end of month because some special case
  ## start date: 2020-01-31 ; end date 2020-06-30
  ## the minimum n month must be 5
  ## another special case of Feb must consider (test case 15)

  if start_date.day > target_date.day && !((start_date == start_date.end_of_month || target_date.month == 2) && (target_date == target_date.end_of_month))
    months_diff = months_diff - 1
  end
  puts months_diff # it will show the minimum whole n month

  # the target_date will between inside
  # (start_date + months_diff.months) <= target_date < (start_date + (months_diff + 1).months) 
  (start_date + months_diff.months)..(start_date + (months_diff + 1).months) 
end

The Test Cases:

## test case 1
## 6/15 - 7/15  => n = 5
calc_diff(Date.parse('2020-01-15'), Date.parse('2020-06-19'))

## test case 2
## 7/15 - 8/15  => n = 6
calc_diff(Date.parse('2020-01-15'), Date.parse('2020-07-15')) 

## test case 3
## 5/15 - 6/15  => n = 4
calc_diff(Date.parse('2020-01-15'), Date.parse('2020-06-01')) 

## test case 4 (special case)
## 6/30 - 7/31  => n = 5
calc_diff(Date.parse('2020-01-31'), Date.parse('2020-06-30'))

## test case 5
## 7/30 - 8/30  => n = 4
calc_diff(Date.parse('2020-04-30'), Date.parse('2020-07-31'))

## test case 6
## 6/30 - 7/30  => n = 2
calc_diff(Date.parse('2020-04-30'), Date.parse('2020-06-30'))  

## test case 7
## 5/31 - 6/30  => n = 4
calc_diff(Date.parse('2020-01-31'), Date.parse('2020-05-31'))

## test case 8
## 2/29 - 3/31  => n = 1
calc_diff(Date.parse('2020-01-31'), Date.parse('2020-02-29'))

## test case 9
## 6/29 - 7/29  => n = 4
calc_diff(Date.parse('2020-02-29'), Date.parse('2020-06-30'))

## test case 10
## 7/29 - 8/29  => n = 5
calc_diff(Date.parse('2020-02-29'), Date.parse('2020-07-31'))

## test case 11
## 1/31 - 2/29  => n = 0
calc_diff(Date.parse('2020-01-31'), Date.parse('2020-02-28'))

## test case 12
## 2/29 - 3/31  => n = 1
calc_diff(Date.parse('2020-01-31'), Date.parse('2020-03-01'))

## test case 13
## 1/17 - 2/17  => n = 0
calc_diff(Date.parse('2020-01-17'), Date.parse('2020-01-17'))

## test case 14
## 1/17 - 2/17  => n = 0
calc_diff(Date.parse('2020-01-17'), Date.parse('2020-01-18'))

## test case 15 (special case)
## 1/30 - 2/29  => n = 1
calc_diff(Date.parse('2019-12-30'), Date.parse('2020-02-28'))

## test case 16
## 2/29 - 3/30  => n = 2
calc_diff(Date.parse('2019-12-30'), Date.parse('2020-02-29'))

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