Selected Intervals of dates in MySQL - mysql

Question

I am working with MySQL. I am trying to get the nights of a booking that belong to each interval in a group of intervals of dates. But there are some intervals that are preferred over others and therefore I will take as many nights for the preferred intervals as possible and fill the gaps with the **not preferred interval **. To illustrate this I will show it here:
Given the dates:
check in => 2016-01-16
check out => 2016-02-08
total nights => 24
Preferred | date_from | date_to | Nights
----------------------------------------------------
1 | 2016-01-15 | 2016-01-17 | 2
1 | 2016-02-03 | 2016-02-10 | 6
1 | 2016-01-20 | 2016-01-25 | 6
0 | 2016-01-20 | 2016-01-31 | 2 (2016-01-26 and 2016-01-31 because the other nights are covered by a preferred period)
1 | 2016-01-27 | 2016-01-30 | 4
0 | 2016-01-15 | 2016-01-17 | 0 (these dates are covered by a the first interval which is a preferred interval )
0 | 2016-02-01 | 2016-02-10 | 2 (just 2016-02-01 and 2016-02-02 because 03 - 08 are covered by the second interval which is a preferred interval)
0 | 2016-01-18 | 2016-01-19 | 2
How can I achieve this in MySQL?

Answer 1

assuming you have a table with columns Preferred,date_from,date_to and you're just trying to calculate # of nights.
You can try this query.
SET #checkin = '2016-01-16';
SET #checkout = '2016-02-08';
SELECT T0.preferred,T0.date_from,T0.date_to,IFNULL(NIGHTS.nights,0) as Nights
FROM YourTable T0
LEFT JOIN
(SELECT T1.preferred,T1.date_from,T1.date_to,COUNT(*) AS Nights
FROM YourTable AS T1
INNER JOIN
(SELECT (#checkin + INTERVAL n DAY) as singleday
FROM numbers
WHERE (#checkin + INTERVAL n DAY) <= #checkout)DAYS1
ON DAYS1.singleday BETWEEN T1.date_from AND T1.date_to
WHERE T1.preferred = 1
OR NOT EXISTS
(SELECT 1
FROM YourTable AS T
WHERE T.preferred = 1
AND DAYS1.singleday BETWEEN T.date_from AND T.date_to
)
GROUP BY T1.preferred,T1.date_from,T1.date_to
)NIGHTS
ON T0.preferred = NIGHTS.preferred
AND T0.date_from = NIGHTS.date_from
AND T0.date_to = NIGHTS.date_to
WHERE
T0.date_from <= #checkout
AND T0.date_to >= #checkin
;
http://sqlfiddle.com/#!9/d64344/10
you can replace #checkout and #checkin occurrences with your actual checkin and check out times.
and you can replace YourTable occurrences with your actual table name
Oh yeah in the sqlfiddle i have included a table called Numbers with column n that contains numbers from 0 counting upward to whatever maximum number of possible days of stay. You need to create this table as well.
to create table numbers use the below
CREATE TABLE numbers AS
SELECT a.n+b.n+c.n+d.n+e.n+f.n+g.n+h.n+i.n as n
FROM
(SELECT 0 as n UNION SELECT 1)a,
(SELECT 0 as n UNION SELECT 2)b,
(SELECT 0 as n UNION SELECT 4)c,
(SELECT 0 as n UNION SELECT 8)d,
(SELECT 0 as n UNION SELECT 16)e,
(SELECT 0 as n UNION SELECT 32)f,
(SELECT 0 as n UNION SELECT 64)g,
(SELECT 0 as n UNION SELECT 128)h,
(SELECT 0 as n UNION SELECT 256)i;
explaination of the query
1) subquery DAYS1 returns all single dates
from #checkin to #checkout range
2) T1 is Joined with DAYS1 WHERE
preferred is 1 OR that there doesnt exist a preferred row that covers
the DAYS1's dates
3) then we do a COUNT(*) GROUP BY
preferred,date_from,date_to to get count of single days
4) Then we call our result NIGHTS
5) Then T0 is LEFT JOINED with NIGHTS to get even rows that have 0 nights
6) And only return T0 rows that intercept out #checkin/#checkout range.
UPDATE If you table is too large you can try and narrow down your subqueries with only rows you're interested in like this
SET #checkin = '2016-01-16';
SET #checkout = '2016-02-08';
SELECT T0.preferred,T0.date_from,T0.date_to,IFNULL(NIGHTS.nights,0) as Nights
FROM (SELECT * FROM YourTable WHERE date_from <= #checkout AND date_to >= #checkin) T0
LEFT JOIN
(SELECT T1.preferred,T1.date_from,T1.date_to,COUNT(*) AS Nights
FROM (SELECT * FROM YourTable WHERE date_from <= #checkout AND date_to >= #checkin) AS T1
INNER JOIN
(SELECT (#checkin + INTERVAL n DAY) as singleday
FROM numbers
WHERE (#checkin + INTERVAL n DAY) <= #checkout)DAYS1
ON DAYS1.singleday BETWEEN T1.date_from AND T1.date_to
WHERE T1.preferred = 1
OR NOT EXISTS
(SELECT 1
FROM (SELECT * FROM YourTable WHERE date_from <= #checkout AND date_to >= #checkin) AS T
WHERE T.preferred = 1
AND DAYS1.singleday BETWEEN T.date_from AND T.date_to
)
GROUP BY T1.preferred,T1.date_from,T1.date_to
)NIGHTS
ON T0.preferred = NIGHTS.preferred
AND T0.date_from = NIGHTS.date_from
AND T0.date_to = NIGHTS.date_to
;