Use MySQL to determine if today is a user's birthday

Question

I have all users' birthdays stored as a unix timestamp and am wanting to send out e-mails each day to users that have a birthday that day. I need to make a MySQL query that will get all of the rows that contain a birthday on today's date. It seems like this should be fairly simple, but maybe I am just overcomplicating it.

Answer 1

set @now=now();
select * from user where (month(birthday) = month(@now) and day(birthday) = day(@now)) or
  (month(birthday) = 2 and day(birthday) = 29 and month(@now) = 2 and day(@now) = 28 and
  month(date_add(@now, interval 1 day)) = 3);

Answer 2

This should work:

   SELECT * 
      FROM USERS
      WHERE 
         DATE_FORMAT(FROM_UNIXTIME(birthDate),'%m-%d') = DATE_FORMAT(NOW(),'%m-%d')

Answer 3

Couldn't you just select all rows that matched the current day's date? You could also use the FROM_UNIXTIME() function to convert from unix timestamp to Date:

mysql> SELECT FROM_UNIXTIME(1196440219); -> '2007-11-30 10:30:19'

This is documented from http://dev.mysql.com/doc/refman/5.1/en/date-and-time-functions.html#function_from-unixtime

Answer 4

The answer below doesn't actually work. It doesn't take into account the fact that a year is 365.24 (leap days now and then) days long, so the actual comparison against the users birthdate is complicated to say the least. I'm leaving it for historical reasons.

The other answers should work but if you want a slight optimization, say if there are many many rows, you are probably better off expressing the query directly in timestamp seconds. You can use the relations (slightly involved because of taking timezone into account):

today_starts = UNIX_TIMESTAMP(NOW()) - TIMESTAMPDIFF(SECOND, DATE(NOW()), NOW())
today_ends = today_starts + 86400

and then select records where the timestamp is between those values.

Answer 5

I took Saggi Malachi's answer and extended to include a birthday on 29th February into 28th February date, if in that year there is no such day.

SELECT * 
      FROM USERS
      WHERE 
         DATE_FORMAT(FROM_UNIXTIME(birthDate),'%m-%d') = DATE_FORMAT(NOW(),'%m-%d')
UNION
SELECT * 
      FROM USERS
      WHERE 
         DATE_FORMAT(NOW(),'%Y')%4 != 0 AND DATE_FORMAT(NOW(),'%m-%d')='02-28' and DATE_FORMAT(FROM_UNIXTIME(birthDate),'%m-%d') = '02-29'

Answer 6

Here is an answer that property takes into account leap-years and will always give you the users whose birthday is on the 29th of February at the same time as those on the 1st of March.

SELECT * 
  FROM USERS
  WHERE 
     DATE_FORMAT(FROM_UNIXTIME(birthDate),'%m-%d') = DATE_FORMAT(NOW(),'%m-%d')
     OR (
            (
                DATE_FORMAT(NOW(),'%Y') % 4 <> 0
                OR (
                        DATE_FORMAT(NOW(),'%Y') % 100 = 0
                        AND DATE_FORMAT(NOW(),'%Y') % 400 <> 0
                    )
            )
            AND DATE_FORMAT(NOW(),'%m-%d') = '03-01'
            AND DATE_FORMAT(FROM_UNIXTIME(birthDate),'%m-%d') = '02-29'
        )

Answer 7

Since this gets more and more to be a code-golf question, here's my approach on solving this including taking care of the leap years:

select * 
from user
where (date_format(from_unixtime(birthday),"%m-%d") = date_format(now(),"%m-%d"))
   or (date_format(from_unixtime(birthday),"%m-%d") = '02-29'
       and date_format('%m') = '02' 
       and last_day(now()) = date(now())
      );

Explanation: The first where clause checks if somebody's birthday is today. The second makes sure to only select those whose birthday is on Feb 29th only if the current day equals the last day of February.

Examples:

SELECT last_day('2009-02-01'); -- gives '2009-02-28'
SELECT last_day('2000-02-01'); -- gives '2009-02-29'
SELECT last_day('2100-02-01'); -- gives '2100-02-28'

Answer 8

Here's my contribution

SELECT
  DAYOFYEAR(CURRENT_DATE)-(dayofyear(date_format(CURRENT_DATE,'%Y-03-01'))-60)=
  DAYOFYEAR(the_birthday)-(dayofyear(date_format(the_birthday,'%Y-03-01'))-60)
FROM
   the_table

The bits '(dayofyear(date_format(current_date,'%Y-03-01'))-60)' returns 1 on leap years since march 1st will be dayofyear number 61, and 0 on normal years.

From here it's just a matter of substracting that extra day to the "is-it-my-birthday"-calculation.