Sehr ähnlich der Antwort von Lalit, aber Sie können eine genaue Anzahl von Tagen erhalten, ohne 30 Tage pro Monat anzunehmen, indem Sie add_months verwenden Anpassung um die gesamte ganzmonatige Differenz:
select sysdate,
hiredate,
trunc(months_between(sysdate,hiredate) / 12) as years,
trunc(months_between(sysdate,hiredate) -
(trunc(months_between(sysdate,hiredate) / 12) * 12)) as months,
trunc(sysdate)
- add_months(hiredate, trunc(months_between(sysdate,hiredate))) as days
from emp;
SYSDATE HIREDATE YEARS MONTHS DAYS
---------- ---------- ---------- ---------- ----------
2015-10-26 1980-12-17 34 10 9
2015-10-26 1981-02-20 34 8 6
2015-10-26 1981-02-22 34 8 4
2015-10-26 1981-04-02 34 6 24
2015-10-26 1981-09-28 34 0 28
2015-10-26 1981-05-01 34 5 25
2015-10-26 1981-06-09 34 4 17
2015-10-26 1982-12-09 32 10 17
2015-10-26 1981-11-17 33 11 9
2015-10-26 1981-09-08 34 1 18
2015-10-26 1983-01-12 32 9 14
2015-10-26 1981-12-03 33 10 23
2015-10-26 1981-12-03 33 10 23
2015-10-26 1982-01-23 33 9 3
Sie können dies überprüfen, indem Sie die Berechnung umkehren:
with tmp as (
select trunc(sysdate) as today,
hiredate,
trunc(months_between(sysdate,hiredate) / 12) as years,
trunc(months_between(sysdate,hiredate) -
(trunc(months_between(sysdate,hiredate) / 12) * 12)) as months,
trunc(sysdate)
- add_months(hiredate, trunc(months_between(sysdate,hiredate))) as days
from emp
)
select * from tmp
where today != add_months(hiredate, (12 * years) + months) + days;
no rows selected
