Question

In MySQL, I can do something like:

SELECT DATE_ADD(‘2010-07-02’, INTERVAL 1 MONTH)

And that returns:

2010-08-02

That’s great… now is there a MySQL function that I can use to find what the date would be one month in advance on the same day. For example, 7/2/2010 falls on the first Friday of July, 2010. Is there an easy way to find what the first Friday of August, 2010 is with an SQL statement?

Answer 1

It got a bit complicated, but here goes:

SELECT IF(MONTH(DATE_ADD('2010-07-02', INTERVAL 28 DAY)) = MONTH('2010-07-02'), 
  DATE_ADD('2010-07-02', INTERVAL 35 DAY),
  DATE_ADD('2010-07-02', INTERVAL 28 DAY));

Rationale: If you add 4 weeks or 5 weeks, you’re still on the same day; since all months are between 28 and 35 days long, if 4 weeks later is still the same month, add another week.

UPDATE: Umm, I did not think this through very well – it works for first X in month, but necessarily for 2nd, 3rd… (i.e. 3rd X in month might return a 2nd X next month). Try #2:

SELECT IF(
  CEIL(DAY(DATE_ADD('2010-07-02', INTERVAL 35 DAY)) / 7) = CEIL(DAY('2010-07-02') / 7), 
  DATE_ADD('2010-07-02', INTERVAL 35 DAY),
  DATE_ADD('2010-07-02', INTERVAL 28 DAY));

Rationale: CEIL(DAY(x) / 7) is x’s week number. If it’s different when you add 5 weeks, then add 4 weeks.

UPDATE 2: LOL, I suck today, I should really think before I post… Week is usually defined as Mon-Sun, or Sun-Mon, not as from whatever started the month till 6 days later. To compensate for this:

SELECT IF(
    CEIL((DAY(DATE_ADD('2010-07-02', INTERVAL 28 DAY)) - DAYOFWEEK('2010-07-02')) / 7)
    = CEIL((DAY('2010-07-02') - DAYOFWEEK('2010-07-02')) / 7), 
  DATE_ADD('2010-07-02', INTERVAL 28 DAY),
  DATE_ADD('2010-07-02', INTERVAL 35 DAY));