Calculate exact date difference in years using SQL - 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 do is a DATEDIFF but the year needs to be exact (i.e., 4.6 years instead of rounding up to five years.
The following is my script:
select *, DATEDIFF (yy, Begin_date, GETDATE()) AS 'Age in Years'
from Report_Stage;
The 'Age_In_Years' column is being rounded. How do I get the exact date in years?

Answer 1

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 2

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

Answer 3

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 4

You want the years difference, but reduced by 1 when the "day of the year" of the future date is less than that of the past date. So like this:
SELECT *
,DATEDIFF(YEAR, [Begin_date], [End_Date])
+ CASE WHEN CAST(DATENAME(DAYOFYEAR, [End_Date]) AS INT)
>= CAST(DATENAME(DAYOFYEAR, [Begin_date]) AS INT)
THEN 0 ELSE -1 END
AS 'Age in Years'
from [myTable];

Answer 5

For me I calculate the difference in days
Declare #startDate datetime
Declare #endDate datetime
Declare #diff int
select #diff=datediff(day,#startDate,#endDate)
if (#diff>=365) then select '1Year'
if (#diff>=730) then select '2Years'
-----etc

Answer 6

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