Oracle Age calculation from Date of birth and Today

I want to calculate the current Age from Date of Birth in my Oracle function.

W开发者_如何学Pythonhat I am using is (Today-Dob)/30/12, but this is not accurate as some months have 31 days.

I need to get the correct age with the maximum precision. How can I do that?

SQL> select trunc(months_between(sysdate,dob)/12) year,
  2         trunc(mod(months_between(sysdate,dob),12)) month,
  3         trunc(sysdate-add_months(dob,trunc(months_between(sysdate,dob)/12)*12+trunc(mod(months_between(sysdate,dob),12)))) day
  4  from (Select to_date('15122000','DDMMYYYY') dob from dual);

      YEAR      MONTH        DAY
---------- ---------- ----------
         9          5         26

SQL>

For business logic I usually find a decimal number (in years) is useful:

select months_between(TRUNC(sysdate),
                      to_date('15-Dec-2000','DD-MON-YYYY')
                     )/12
as age from dual;

       AGE
----------
9.48924731

SELECT 
TRUNC((SYSDATE - TO_DATE(DOB, 'YYYY-MM-DD'))/ 365.25) AS AGE_TODAY FROM DUAL;

This is easy and straight to the point.

Age (full years) of the Person:

SELECT
  TRUNC(months_between(sysdate, per.DATE_OF_BIRTH) / 12) AS "Age"
FROM PD_PERSONS per

Or how about this?

with some_birthdays as
( 
    select date '1968-06-09' d from dual union all
    select date '1970-06-10' from dual union all
    select date '1972-06-11' from dual union all
    select date '1974-12-11' from dual union all
    select date '1976-09-17' from dual
)
select trunc(sysdate) today
, d birth_date
, floor(months_between(trunc(sysdate),d)/12) age
from some_birthdays;

And an alternative without using any arithmetic and numbers (although there is nothing wrong with that):

SQL> with some_birthdays as
  2  ( select date '1968-06-09' d from dual union all
  3    select date '1970-06-10' from dual union all
  4    select date '1972-06-11' from dual union all
  5    select date '1974-12-11' from dual union all
  6    select date '1976-09-17' from dual
  7  )
  8  select trunc(sysdate) today
  9       , d birth_date
 10       , extract(year from numtoyminterval(months_between(trunc(sysdate),d),'month')) age
 11    from some_birthdays
 12  /

TODAY               BIRTH_DATE                 AGE
------------------- ------------------- ----------
10-06-2010 00:00:00 09-06-1968 00:00:00         42
10-06-2010 00:00:00 10-06-1970 00:00:00         40
10-06-2010 00:00:00 11-06-1972 00:00:00         37
10-06-2010 00:00:00 11-12-1974 00:00:00         35
10-06-2010 00:00:00 17-09-1976 00:00:00         33

5 rows selected.

You can try

SELECT ROUND((SYSDATE - TO_DATE('12-MAY-16'))/365.25, 5) AS AGE from DUAL;

You can configure ROUND to show as many decimal places as you wish.

Placing the date in decimal format like aforementioned helps with calculations of age groups, etc.

This is just a contrived example. In real world scenarios, you wouldn't be converting strings to date using TO_DATE.

However, if you have date of birth in date format, you can subtract two dates safely.

This seems considerably easier than what anyone else has suggested

select sysdate-to_date('30-jul-1977') from dual; 

Suppose that you want to have the age (number of years only, a fixed number) of someone born on June 4, 1996, execute this command :

SELECT TRUNC(TO_NUMBER(SYSDATE - TO_DATE('04-06-1996')) / 365.25) AS AGE FROM DUAL;

Result : (Executed May 28, 2019)

       AGE
----------
        22

Explanation :

• SYSDATE : Get system's (OS) actual date.
• TO_DATE('04-06-1996') : Convert VARCHAR (string) birthdate into DATE (SQL type).
• TO_NUMBER(...) : Convert a date to NUMBER (SQL type)
• Devide per 365.25 : To have a bissextile year every four years (4 * 0.25 = 1 more day).
• Trunc(...) : Retrieve the entire part only from a number.

SQL>select to_char(to_date('19-11-2017','dd-mm-yyyy'),'yyyy') -  to_char(to_date('10-07-1986','dd-mm-yyyy'),'yyyy') year,
to_char(to_date('19-11-2017','dd-mm-yyyy'),'mm') -  to_char(to_date('10-07-1986','dd-mm-yyyy'),'mm') month,
to_char(to_date('19-11-2017','dd-mm-yyyy'),'dd') -  to_char(to_date('10-07-1986','dd-mm-yyyy'),'dd') day from dual;

      YEAR      MONTH        DAY
---------- ---------- ----------
        31          4          9

    SELECT FLOOR(MONTHS_BETWEEN(TRUNC(SYSDATE, 'DD'), dob)/12)
    FROM dual;

You can try below method,

SELECT EXTRACT(YEAR FROM APP_SUBMITTED_DATE)-EXTRACT(YEAR FROM BIRTH_DATE) FROM SOME_TABLE;

It will compare years and give age accordingly.
You can also use SYSDATE instead of APP_SUBMITTED_DATE.

Regards.

Select trunc(to_number(sysdate - to_date(dob) /365.25)) as Age from desingh

AGE

26 25 31

select (extract(year from current_date)-extract(year from Date_of_birth)) as Age from table_name;`

age=current_year - birth_year;

extract(year/month/date from date) //oracle function for extracting values from date

select (SYSDATE-DOB)/365 "Age" from dual