How to get age in years,months and days using Oracle

问题

I'm trying to print for each person its age using this format :

E.g : 19 years , 8 months , 13 days.

I've googled a lot and I've noticed that there is a specific function to calculate the difference between dates DATEDIFF.

However this function does not exist in SQL*Plus , so I went on trying using MONTHS_BETWEEN() and some operators.

My attempt:

SELECT name , ' ' || 
    FLOOR(MONTHS_BETWEEN(to_date(SYSDATE),to_date(date_of_birth))/12)||' years ' ||  
    FLOOR(MOD(MONTHS_BETWEEN(to_date(SYSDATE),to_date(date_of_birth)),12)) || ' months ' || 
    FLOOR(MOD(MOD(MONTHS_BETWEEN(to_date(SYSDATE),to_date(date_of_birth)),12),4))|| ' days ' AS "Age"
FROM persons;

My issue relies on getting the days. I don't know how should I calculate the days , using this function ('tried dividing by 4 , or 30); I'm thinking my logic is bad but I can't figure it out , any ideas ?

回答1:

Very similar to Lalit's answer, but you can get an accurate number of days without assuming 30 days per month, by using add_months to adjust by the total whole-month difference:

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

You can verify by reversing the calculation:

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

回答2:

Getting the age in terms of YEARS and MONTHS is easy, but the tricky part is the the DAYS.

If you can fix the days in a month, you could get the number of days in the same SQL. For example, using the standard SCOTT.EMP table and assuming every month has 30 days:

SQL> SELECT SYSDATE,
  2        hiredate,
  3        TRUNC(months_between(SYSDATE,hiredate)/12) years,
  4        TRUNC(months_between(SYSDATE,hiredate)  -
  5        (TRUNC(months_between(SYSDATE,hiredate)/12)*12)) months,
  6        TRUNC((months_between(SYSDATE,hiredate) -
  7        TRUNC(months_between(SYSDATE,hiredate)))*30) days
  8  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         23
2015-10-26 1981-09-28         34          0         28
2015-10-26 1981-05-01         34          5         24
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         22
2015-10-26 1981-12-03         33         10         22
2015-10-26 1982-01-23         33          9          3

14 rows selected.

But, be aware not every month has 30 days. So, you cannot have the accuracy with number of days.

UPDATE

I missed the total whole-month difference which @Alex Poole has explained in his accepted answer. I will let this answer for future readers to understand the part that was missed about calculating the number of days.

Modify this:

TRUNC((months_between(SYSDATE,hiredate) -       
TRUNC(months_between(SYSDATE,hiredate)))*30) days

With this:

TRUNC(SYSDATE) - add_months(hiredate, TRUNC(months_between(sysdate,hiredate)))

回答3:

Syntax:

SELECT 
  CONCAT(
    TIMESTAMPDIFF(YEAR, ?, NOW()),
    ' Years,',
    TIMESTAMPDIFF(MONTH, ?, NOW()) % 12,
    ' Months,',
    FLOOR(TIMESTAMPDIFF(DAY, ?, NOW()) % 30.4375),
    ' Days'
  ) AS age 
FROM
  DUAL

Input: replace '?' with Date of Birth. For Example, '1994-07-08'
Output: This query will return age in 'X' Years 'Y' Months and 'Z' days.

回答4:

Another simplified way for getting months would be-

TRUNC(MOD(months_between(sysdate,hiredate),12)) AS months

来源：https://stackoverflow.com/questions/33343596/how-to-get-age-in-years-months-and-days-using-oracle

标签

sql

Oracle

date-arithmetic