Find date gaps with mysql

Question

I would like to extract records with having an empty bookingId and get the maximum unbooked days back (from the first free day). The expected result should be:

id = 1, 2013-08-03, 7 days free
id = 1, 2013-08-24, 7 days free
id = 2, 2013-08-07, 10 days free
id = 2, 2013-08-24, 7 days free

The best thing would be, If I can also query for a free timeslot: e.g. query for 1,2,3,4,5,6,7..14.. free days. This is an example of my source data:

id      bookingDate    bookingId
--------------------------------
1        2013-08-03        0
1        2013-08-04        0
1        2013-08-05        0
1        2013-08-06        0
1        2013-08-07        0
1        2013-08-08        0
1        2013-08-09        0
1        2013-08-10        112
1        2013-08-11        112
1        2013-08-12        112
1        2013-08-13        112
1        2013-08-14        112
1        2013-08-15        112
1        2013-08-16        112
1        2013-08-17        112
1        2013-08-18        112
1        2013-08-19        112
1        2013-08-20        112
1        2013-08-21        112
1        2013-08-22        112
1        2013-08-23        112
1        2013-08-24        0
1        2013-08-25        0
1        2013-08-26        0
1        2013-08-27        0
1        2013-08-28        0
1        2013-08-29        0
1        2013-08-30        0
1        2013-08-31        0
2        2013-08-03        78
2        2013-08-04        78
2        2013-08-05        78
2        2013-08-06        78
2        2013-08-07        0
2        2013-08-08        0
2        2013-08-09        0
2        2013-08-10        0
2        2013-08-11        0
2        2013-08-12        0
2        2013-08-13        0
2        2013-08-14        0
2        2013-08-15        0
2        2013-08-16        0
2        2013-08-17        39
2        2013-08-18        39
2        2013-08-19        39
2        2013-08-20        39
2        2013-08-21        39
2        2013-08-22        39
2        2013-08-23        39
2        2013-08-24        0
2        2013-08-25        0
2        2013-08-26        0
2        2013-08-27        0
2        2013-08-28        0
2        2013-08-29        0
2        2013-08-30        0
2        2013-08-31        0

If anyone has an good idea for a better data structure, I can try to implement. The database is still under construction :-)

Edit:

CREATE TABLE IF NOT EXISTS `pricesBookings` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `baseId` int(11) NOT NULL,
  `bookingDate` date NOT NULL,
  `bookingId` int(11) NOT NULL,
  `price` decimal(10,2) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `baseId` (`baseId`,`bookingDate`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;

Answer 1

Pls try this...

select
  concat_ws(',',(concat("ID=",id)),
  min(startDate),
 (concat((count(*) - (endDate is null))," Days Free"))) as result
from (
  select
    pb1.id,
    pb1.bookingDate startDate,
    min(pb2.bookingDate) endDate
  from
    pricesBookings pb1 left join pricesBookings pb2
    on pb1.id=pb2.id
       and pb2.price>0
       and pb2.bookingDate>pb1.bookingDate
  where
    pb1.price=0
  group by
    pb1.id,
    pb1.bookingDate
) s
group by id, endDate
  order by id, startDateselect
  concat_ws(',',(concat("ID=",id)),
  min(startDate),
 (concat((count(*) - (endDate is null))," Days Free"))) as result
from (
  select
    pb1.id,
    pb1.bookingDate startDate,
    min(pb2.bookingDate) endDate
  from
    pricesBookings pb1 left join pricesBookings pb2
    on pb1.id=pb2.id
       and pb2.price>0
       and pb2.bookingDate>pb1.bookingDate
  where
    pb1.price=0
  group by
    pb1.id,
    pb1.bookingDate
) s
group by id, endDate
  order by id, startDate

Answer 2

This should give the correct result:

select
  id,
  min(startDate) as startFreeDate,
  count(*) - (endDate is null) numFreeDays
from (
  select
    pb1.id,
    pb1.bookingDate startDate,
    min(pb2.bookingDate) endDate
  from
    pricesBookings pb1 left join pricesBookings pb2
    on pb1.id=pb2.id
       and pb2.price>0
       and pb2.bookingDate>pb1.bookingDate
  where
    pb1.price=0
  group by
    pb1.id,
    pb1.bookingDate
) s
group by id, endDate
order by id, startDate

see it here.

If you need to search for all free slots of, for example, 14 days, you can add HAVING:

group by id, endDate
having count(*) - (endDate is null) >= 14
order by id, startDate

Answer 3

Had a play with this. I might be missing something obvious but I can't see an easy way to do this with a single statement.

But I have come up with this nasty way of doing it.

SELECT z.baseid, z.bookingdate, 
CASE 
    WHEN j.id IS NOT NULL THEN '11+ days free'
    WHEN i.id IS NOT NULL THEN '10 days free'
    WHEN h.id IS NOT NULL THEN '9 days free'
    WHEN g.id IS NOT NULL THEN '8 days free'
    WHEN f.id IS NOT NULL THEN '7 days free'
    WHEN e.id IS NOT NULL THEN '6 days free'
    WHEN d.id IS NOT NULL THEN '5 days free'
    WHEN c.id IS NOT NULL THEN '4 days free'
    WHEN b.id IS NOT NULL THEN '3 days free'
    WHEN a.id IS NOT NULL THEN '2 days free'
    ELSE '1 day free'
END AS DaysFree
FROM pricesbookings z
INNER JOIN pricesbookings y
ON z.baseid = y.baseid AND z.bookingid = 0 AND y.bookingid != 0 AND DATE_ADD(y.bookingdate, INTERVAL 1 DAY) = z.bookingdate
LEFT JOIN pricesbookings a ON z.baseid = a.baseid AND z.bookingid = 0 AND a.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 1 DAY) = a.bookingdate
LEFT OUTER JOIN pricesbookings b ON a.baseid = b.baseid AND b.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 2 DAY) = b.bookingdate
LEFT OUTER JOIN pricesbookings c ON b.baseid = c.baseid AND c.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 3 DAY) = c.bookingdate
LEFT OUTER JOIN pricesbookings d ON c.baseid = d.baseid AND d.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 4 DAY) = d.bookingdate
LEFT OUTER JOIN pricesbookings e ON d.baseid = e.baseid AND e.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 5 DAY) = e.bookingdate
LEFT OUTER JOIN pricesbookings f ON e.baseid = f.baseid AND f.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 6 DAY) = f.bookingdate
LEFT OUTER JOIN pricesbookings g ON f.baseid = g.baseid AND g.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 7 DAY) = g.bookingdate
LEFT OUTER JOIN pricesbookings h ON g.baseid = h.baseid AND h.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 8 DAY) = h.bookingdate
LEFT OUTER JOIN pricesbookings i ON h.baseid = i.baseid AND i.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 9 DAY) = i.bookingdate
LEFT OUTER JOIN pricesbookings j ON i.baseid = j.baseid AND j.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 10 DAY) = j.bookingdate
ORDER BY z.baseid, z.bookingdate

This only counts up to 11 or more days (easy to expand if you need to, but does need tha max number to be known in advance), and probably hideously inefficient.

Basically the table alias z is the first day, which is joined against table alias y to check that the previous day was booked. Then LEFT JOINs against a load more copies of the table each with an extra day added to the date. Then uses a CASE statement to check which is the largest one found to give you the number of days free.

Works, but your database might not appreciate it!