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Come ottenere l'età in anni, mesi e giorni utilizzando Oracle

Molto simile alla risposta di Lalit, ma puoi ottenere un numero preciso di giorni senza assumere 30 giorni al mese, utilizzando add_months per adeguare la differenza di un mese intero totale:

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

Puoi verificare invertendo il calcolo:

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