MySQL日期查找器从今天开始

Question

I have a table of users which has a date field for their birthday:

buddy_auto_id    int(11) PK

user_id          varchar(128)

buddy_user_id    varchar(128)

buddy_name       varchar(128)

buddy_bday       date

buddyuser_id     varchar(20)

active           enum('Yes','No')

requestsentby    int(11)

whenrequested    timestamp

I'm trying to find the 3 users whose birthdays fall soonest compared to todays date and then display the number of days until their birthday ordered by soonest first.

Is this possible within a SQL query or do I have to pull it out and let PHP do the equation?

Many thanks

Answer 1

We first need to calculate the next birthday, then order by that value:

select *,
    buddy_bday + interval 
        if(
              month(buddy_bday) < month(now()) or 
              (month(buddy_bday) = month(now()) and day(buddy_bday) < day(now())),

              year(now())+1,
              year(now())
        ) - year(buddy_bday) year as next_bday
from buddies order by next_bday - date(now());

The long if statement figures out whether the buddy already had his/her birthday this year.

Answer 2

This should be possible with SQL. You need to compare the current date with a birthday expressed as from this year or next year, depending on whether we've past it in the current year. Once you have this "next birthday" date, use DATEDIFF function to determine number of days distant from current date.

SELECT *,
    DATEDIFF(
        # determine date of next birthday by adding age in years to birthday year
        DATE_ADD(buddy_bday, INTERVAL
            YEAR(CURDATE())-YEAR(buddy_bday)
            # add a year if we celebrated birthday already this year
            +(MONTH(buddy_bday)<MONTH(CURDATE()) OR (MONTH(buddy_bday)=MONTH(CURDATE()) AND DAY(buddy_bday) < DAY(CURDATE())))
        YEAR),
        CURDATE())
    AS days_to_next_bday
FROM user_table
ORDER BY days_to_next_bday
LIMIT 3;

Answer 3

You should use a DAYOFYEAR function. Try this query -

SELECT buddy_auto_id , buddy_bday FROM table_name
  WHERE DAYOFYEAR(buddy_bday) - DAYOFYEAR(NOW()) > 0
  ORDER BY DAYOFYEAR(buddy_bday) - DAYOFYEAR(NOW())
  LIMIT 3;

So, this query works only for current year.

---

EDITED query 2:

This one works for all dates.

CREATE TABLE birtdays(
  buddy_auto_id INT(11) NOT NULL AUTO_INCREMENT,
  buddy_bday DATE DEFAULT NULL,
  PRIMARY KEY (buddy_auto_id)
);

INSERT INTO birtdays VALUES 
  (1, '2011-10-04'),
  (2, '2011-03-01'),
  (3, '2011-11-29'),
  (4, '2011-11-10'),
  (5, '2011-12-29'),
  (6, '2011-11-30'),
  (7, '2011-12-08'),
  (8, '2011-09-17'),
  (9, '2011-12-01'),
  (10, '2011-12-11');

SELECT buddy_auto_id, buddy_bday
  FROM
    birtdays, (SELECT @day_of_year:=DAYOFYEAR(NOW())) t
  ORDER BY
    DAYOFYEAR(buddy_bday + INTERVAL YEAR(NOW()) - YEAR(buddy_bday) YEAR) - @day_of_year
      + IF (DAYOFYEAR(buddy_bday + INTERVAL YEAR(NOW()) - YEAR(buddy_bday) YEAR) - @day_of_year > 0, 0, DAYOFYEAR(STR_TO_DATE(CONCAT(YEAR(NOW()), '-12-31'), '%Y-%m-%d')))
  LIMIT 3;

+---------------+------------+
| buddy_auto_id | buddy_bday |
+---------------+------------+
|             6 | 2011-11-30 |
|             7 | 2011-12-08 |
|            10 | 2012-02-11 |
+---------------+------------+