Calculate exact date difference in years using SQL

Question

I receive reports in which the data is ETL to the DB automatically. I extract and transform some of that data to load it somewhere else. One thing I need to d

Answer 1

I have found a better solution. This makes the assumption that the first date is less than or equal to the second date.

declare @dateTable table (date1 datetime, date2 datetime)
insert into @dateTable 
    select '2017-12-31', '2018-01-02' union
    select '2017-01-03', '2018-01-02' union 
    select '2017-01-02', '2018-01-02' union
    select '2017-01-01', '2018-01-02' union
    select '2016-12-01', '2018-01-02' union
    select '2016-01-03', '2018-01-02' union
    select '2016-01-02', '2018-01-02' union
    select '2016-01-01', '2018-01-02' 
select date1, date2, 
        case when ((DATEPART(year, date1) < DATEPART(year, date2)) and 
                    ((DATEPART(month, date1) <= DATEPART(month, date2)) and 
(DATEPART(day, date1) <= DATEPART(day, date2)) ))
                    then DATEDIFF(year, date1, date2)
            when (DATEPART(year, date1) < DATEPART(year, date2))
                    then DATEDIFF(year, date1, date2) - 1
            when (DATEPART(year, date1) = DATEPART(year, date2))
                    then 0
        end [YearsOfService]
from @dateTable

date1                   date2                   YearsOfService
----------------------- ----------------------- --------------
2016-01-01 00:00:00.000 2018-01-02 00:00:00.000 2
2016-01-02 00:00:00.000 2018-01-02 00:00:00.000 2
2016-01-03 00:00:00.000 2018-01-02 00:00:00.000 1
2016-12-01 00:00:00.000 2018-01-02 00:00:00.000 1
2017-01-01 00:00:00.000 2018-01-02 00:00:00.000 1
2017-01-02 00:00:00.000 2018-01-02 00:00:00.000 1
2017-01-03 00:00:00.000 2018-01-02 00:00:00.000 0
2017-12-31 00:00:00.000 2018-01-02 00:00:00.000 0

Answer 2

I think that division by 365.2425 is not a good way to do this. No division can to this completely accurately (using 365.25 also has issues).

I know the following script calculates an accurate date difference (though might not be the most speedy way):

        declare @d1 datetime ,@d2 datetime
        --set your dates eg: 
        select @d1 = '1901-03-02'
        select @d2 = '2016-03-01'

        select DATEDIFF(yy, @d1, @d2) -
            CASE WHEN MONTH(@d2) < MONTH(@d1) THEN 1
                 WHEN MONTH(@d2) > MONTH(@d1) THEN 0
                 WHEN DAY(@d2) < DAY(@d1) THEN 1
                 ELSE 0 END

         -- = 114 years

For comparison:

         select datediff(day,@d1 ,@d2) / 365.2425
         -- = 115 years => wrong!

You might be able to calculate small ranges with division, but why take a chance??

The following script can help to test yeardiff functions (just swap cast(datediff(day,@d1,@d2) / 365.2425 as int) to whatever the function is):

   declare @d1 datetime set @d1 = '1900-01-01'

   while(@d1 < '2016-01-01')
   begin
    declare @d2 datetime set @d2 = '2016-04-01'

    while(@d2 >= '1900-01-01')
    begin
        if (@d1 <= @d2 and dateadd(YEAR,     cast(datediff(day,@d1,@d2) / 365.2425 as int)      , @d1) > @d2)
        begin
            select 'not a year!!', @d1, @d2, cast(datediff(day,@d1,@d2) / 365.2425 as int)
        end

        set @d2 = dateadd(day,-1,@d2)
    end

    set @d1 = dateadd(day,1,@d1)
  end

Answer 3

All datediff() does is compute the number of period boundaries crossed between two dates. For instance

datediff(yy,'31 Dec 2013','1 Jan 2014')

returns 1.

You'll get a more accurate result if you compute the difference between the two dates in days and divide by the mean length of a calendar year in days over a 400 year span (365.2425):

datediff(day,{start-date},{end-date},) / 365.2425

For instance,

select datediff(day,'1 Jan 2000' ,'18 April 2014') / 365.2425

return 14.29461248 — just round it to the desired precision.

Answer 4

Have you tried getting the difference in months instead and then calculating the years that way? For example 30 months / 12 would be 2.5 years.

Edit: This SQL query contains several approaches to calculate the date difference:

SELECT CONVERT(date, GetDate() - 912) AS calcDate
      ,DATEDIFF(DAY, GetDate() - 912, GetDate()) diffDays
      ,DATEDIFF(DAY, GetDate() - 912, GetDate()) / 365.0 diffDaysCalc
      ,DATEDIFF(MONTH, GetDate() - 912, GetDate()) diffMonths
      ,DATEDIFF(MONTH, GetDate() - 912, GetDate()) / 12.0 diffMonthsCalc
      ,DATEDIFF(YEAR, GetDate() - 912, GetDate()) diffYears