How to get difference between two dates in Years, Months and days in Sql Server code example

Example 1: date difference in number of days sql server

DECLARE 
    @start_dt DATETIME2= '2019-12-31 23:59:59.9999999', 
    @end_dt DATETIME2= '2020-01-01 00:00:00.0000000';

SELECT 
    DATEDIFF(year, @start_dt, @end_dt) diff_in_year, 
    DATEDIFF(quarter, @start_dt, @end_dt) diff_in_quarter, 
    DATEDIFF(month, @start_dt, @end_dt) diff_in_month, 
    DATEDIFF(dayofyear, @start_dt, @end_dt) diff_in_dayofyear, 
    DATEDIFF(day, @start_dt, @end_dt) diff_in_day, 
    DATEDIFF(week, @start_dt, @end_dt) diff_in_week, 
    DATEDIFF(hour, @start_dt, @end_dt) diff_in_hour, 
    DATEDIFF(minute, @start_dt, @end_dt) diff_in_minute, 
    DATEDIFF(second, @start_dt, @end_dt) diff_in_second, 
    DATEDIFF(millisecond, @start_dt, @end_dt) diff_in_millisecond;

Example 2: How to get number of months between 2 dates sql server

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