mysql find date where no row exists for previous day - mysql

Question

I need to select how many days since there is a break in my data. It's easier to show:
Table format:
id (autoincrement), user_id (int), start (datetime), end (datetime)
Example data (times left out as only need days):
1, 5, 2011-12-18, 2011-12-18
2, 5, 2011-12-17, 2011-12-17
3, 5, 2011-12-16, 2011-12-16
4, 5, 2011-12-13, 2011-12-13
As you can see there would be a break between 2011-12-13 and 2011-12-16. Now, I need to be able say:
Using the date 2011-12-18, how many days are there until a break:
2011-12-18: Lowest sequential date = 2011-12-16: Total consecutive days: 3
Probably: DATE_DIFF(2011-12-18, 2011-12-16)
So my problem is, how can I select that 2011-12-16 is the lowest sequential date? Remembering that data applies for particular user_id's.
It's kinda like the example here: http://www.artfulsoftware.com/infotree/queries.php#72 but in the reverse.
I'd like this done in SQL only, no php code
Thanks

Answer 1

SELECT qmin.start, qmax.end, DATE_DIFF( qmax.end, qmin.start ) FROM table AS qmin
LEFT JOIN (
SELECT end FROM table AS t1
LEFT JOIN table AS t2 ON
t2.start > t1.end AND
t2.start < DATE_ADD( t1.end, 1 DAY )
WHERE t1.end >= '2011-12-18' AND t2.start IS NULL
ORDER BY end ASC LIMIT 1
) AS qmax
LEFT JOIN table AS t2 ON
t2.end < qmin.start AND
t2.end > DATE_DIFF( qmin.start, 1 DAY )
WHERE qmin.start <= '2011-12-18' AND t2.start IS NULL
ORDER BY end DESC LIMIT 1
This should work - left joins selects one date which can be in sequence, so max can be fineded out if you take the nearest record without sequential record ( t2.anyfield is null ) , same thing we do with minimal date.
If you can calculate days between in script - do it using unions ( eg 1. row - minimal, 2. row maximal )

Answer 2

Check this,
SELECT DATEDIFF((SELECT MAX(`start`) FROM testtbl WHERE `user_id`=1),
(select a.`start` from testtbl as a
left outer join testtbl as b on a.user_id = b.user_id
AND a.`start` = b.`start` + INTERVAL 1 DAY
where a.user_id=1 AND b.`start` is null
ORDER BY a.`start` desc LIMIT 1))
DATEDIFF() show difference of the Two days, if you want to number of consecutive days add one for that result.

Answer 3

If it's not a beauty contents then you may try something like:
select t.start, t2.start, datediff(t2.start, t.start) + 1 as consecutive_days
from tab t
join tab t2 on t2.start = (select min(start) from (
select c1.*, case when c2.id is null then 1 else 0 end as gap
from tab c1
left join tab c2 on c1.start = adddate(c2.start, -1)
) t4 where t4.start <= t.start and t4.start >= (select max(start) from (
select c1.*, case when c2.id is null then 1 else 0 end as gap
from tab c1
left join tab c2 on c1.start = adddate(c2.start, -1)
) t3 where t3.start <= t.start and t3.gap = 1))
where t.start = '2011-12-18'
Result should be:
start start consecutive_days
2011-12-18 2011-12-16 3