Get difference in years between two dates in MySQL as an integer

Question

I am trying to calculate how old is a person in a database.
Let\'s suppose to have this simple table:

student(id, birth_date);

Where

Answer 1

I have not enough reputation to add comment to an answer by Rat-a-tat-a-tat Ratatouille to improve his code. Here is the better SQL-query:

SELECT IFNULL(TIMESTAMPDIFF(YEAR, birthdate, CURDATE()), YEAR(CURDATE()) - YEAR(birthdate)) AS age

This is better because sometimes "birthdate" may contain only year of birth, while day and month is 0. If TIMESTAMPDIFF() returns NULL, we can find rough age by subtracting the current year from the year of birth.

Answer 2

I'd suggest this:

DATE_FORMAT(NOW(),"%Y")
   -DATE_FORMAT(BirthDate,'%Y')
   -(
     IF(
      DATE_FORMAT(NOW(),"%m-%d") < DATE_FORMAT(BrthDate,'%m-%d'),
      1,
      0
     )
    ) as Age

This should work with leap-years very well ;-)

Answer 3

Try:

SELECT 
  DATE_FORMAT(FROM_DAYS(TO_DAYS(NOW())-TO_DAYS(birth_date)), '%Y')+0
  AS age FROM student;

Answer 4

This works, even taking in account leap years!

select floor((cast(date_format('2016-02-29','%Y%m%d') as int) - cast(date_format('1966-03-01','%Y%m%d') as int)/10000);

Just be sure to keep the floor() as the decimal is not needed

Answer 5

For anyone who comes across this:

another way this can be done is:

SELECT TIMESTAMPDIFF(YEAR, date_of_birth, CURDATE()) AS difference FROM student

For differences in months, replace YEAR with MONTH, and for days replace YEAR with DAY

Hope that helps!

Answer 6

Why not use MySQL's FLOOR() function on the output from your second approach? Any fractions will be dropped, giving you the result you want.