Oracle date difference to get number of years

I'd use months_between, possibly combined with floor:

select floor(months_between(date '2012-10-10', date '2011-10-10') /12) from dual;select floor(months_between(date '2012-10-9' , date '2011-10-10') /12) from dual;

floor makes sure you get down-rounded years. If you want the fractional parts, you obviously want to not use floor.

If you just want the difference in years, there's:

SELECT EXTRACT(YEAR FROM date1) - EXTRACT(YEAR FROM date2) FROM mytable

Or do you want fractional years as well?

SELECT (date1 - date2) / 365.242199 FROM mytable

365.242199 is 1 year in days, according to Google.

I had to implement a year diff function which works similarly to sybase datediff. In that case the real year difference is counted, not the rounded day difference. So if there are two dates separated by one day, the year difference can be 1 (see select datediff(year, '20141231', '20150101')).

If the year diff has to be counted this way then use:

EXTRACT(YEAR FROM date_to) - EXTRACT(YEAR FROM date_from)

Just for the log the (almost) complete datediff function:

CREATE OR REPLACE FUNCTION datediff (datepart IN VARCHAR2, date_from IN DATE, date_to IN DATE)RETURN NUMBERAS  diff NUMBER;BEGIN  diff :=  CASE datepart    WHEN 'day'   THEN TRUNC(date_to,'DD') - TRUNC(date_from, 'DD')    WHEN 'week'  THEN (TRUNC(date_to,'DAY') - TRUNC(date_from, 'DAY')) / 7    WHEN 'month' THEN MONTHS_BETWEEN(TRUNC(date_to, 'MONTH'), TRUNC(date_from, 'MONTH'))    WHEN 'year'  THEN EXTRACT(YEAR FROM date_to) - EXTRACT(YEAR FROM date_from)  END;  RETURN diff;END;";