如何使用Oracle获得年,月,日的年龄

Question

我正在尝试使用以下格式为每个人打印年龄:

例如:19年,8个月,13天.

我已经google了很多,我注意到有一个特定的函数来计算日期之间的差异DATEDIFF.

但是这个功能不存在SQL*Plus,所以我继续尝试使用MONTHS_BETWEEN()和一些运算符.

我的尝试:

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;

我的问题依赖于获得日子.我不知道如何计算天数,使用此功能('尝试除以4或30); 我认为我的逻辑很糟糕,但我无法弄明白,任何想法？

Answer 1

与Lalit的回答非常相似,但您可以通过使用add_months整月差异调整来获得准确的天数而不假设每月30天:

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

您可以通过反转计算来验证:

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