SQL "First relevant day" - mysql

Question

I have a table with opening_hours for restaurants:
SELECT * FROM opening_hours;
+----+---------------+------------+----------+-----+
| id | restaurant_id | start_time | end_time | day |
+----+---------------+------------+----------+-----+
| 1 | 1 | 12:00:00 | 18:00:00 | 1 |
| 2 | 1 | 09:00:00 | 19:00:00 | 4 |
| 3 | 2 | 09:00:00 | 16:00:00 | 4 |
| 4 | 2 | 09:00:00 | 16:00:00 | 5 |
| 5 | 3 | 09:00:00 | 16:00:00 | 4 |
| 6 | 3 | 09:00:00 | 16:00:00 | 5 |
| 7 | 3 | 09:00:00 | 16:00:00 | 1 |
| 8 | 3 | 09:00:00 | 16:00:00 | 6 |
+----+---------------+------------+----------+-----+
http://www.sqlfiddle.com/#!2/eaea09/1
Now I want to fetch the "closest" next day or same day for every restaurant to the current day. For example when the current day is 1 the result would be:
restaurant_id: 1 day: 1
restaurant_id: 2 day: 4
restaurant_id: 3 day: 1
In the case of day 1 I could do this:
SELECT day FROM opening_hours WHERE day >= 1 GROUP BY restaurant_id LIMIT 1
But if today would be 6 that would not work. I would need the query to go get the maximum number of days (7) and if that could not be found it should start trying from 1 again. So the result for day 6 would be in this case:
restaurant_id: 1 day: 1
restaurant_id: 2 day: 4
restaurant_id: 3 day: 6
How could I achieve this with a query?
I'd think it could be something like this in pseudo SQL:
SELECT `day` FROM opening_hours WHERE `day` >= 'today' IF NOT FOUND WHERE `day` >= 1 GROUP BY `restaurant_id` LIMIT 1
edit:
I could run 2 queries, and determine if a match for a restaurant was found in the first. If not, run a second query. But there must be a better way.

Answer 1

Tested, this one works :) you can keep it simple.
Query:
SELECT * FROM (
SELECT
oh.*
FROM
opening_hours oh
ORDER BY restaurant_id,
`day` + IF(`day` < $current_day, 7, 0)
) sq
GROUP BY restaurant_id;
Explanation:
Note though, that this is a bit hacky. To select a column that is not used in the group by and has no aggregate function applied to it, usually isn't allowed, because theoretically it could give you a random row of each group. That's why it's not allowed in most database systems. MySQL is actually the only one I know of, that allows this (if not set otherwise via sql-mode). Like I said, in theory. Practically it's a bit different and if you do an order by in the subquery, MySQL will always give you the minimum or maximum value (depending on the sort order).
Tests:
Desired result with current day = 1:
root#VM:playground > SELECT * FROM (
-> SELECT
-> oh.*
-> FROM
-> opening_hours oh
-> ORDER BY restaurant_id,
-> `day` + IF(`day` < 1, 7, 0)
-> ) sq
-> GROUP BY restaurant_id;
+----+---------------+------------+----------+-----+
| id | restaurant_id | start_time | end_time | day |
+----+---------------+------------+----------+-----+
| 1 | 1 | 12:00:00 | 18:00:00 | 1 |
| 3 | 2 | 09:00:00 | 16:00:00 | 4 |
| 7 | 3 | 09:00:00 | 16:00:00 | 1 |
+----+---------------+------------+----------+-----+
3 rows in set (0.00 sec)
Desired result with current day = 6:
root#VM:playground > SELECT * FROM (
-> SELECT
-> oh.*
-> FROM
-> opening_hours oh
-> ORDER BY restaurant_id,
-> `day` + IF(`day` < 6, 7, 0)
-> ) sq
-> GROUP BY restaurant_id;
+----+---------------+------------+----------+-----+
| id | restaurant_id | start_time | end_time | day |
+----+---------------+------------+----------+-----+
| 1 | 1 | 12:00:00 | 18:00:00 | 1 |
| 3 | 2 | 09:00:00 | 16:00:00 | 4 |
| 8 | 3 | 09:00:00 | 16:00:00 | 6 |
+----+---------------+------------+----------+-----+
3 rows in set (0.00 sec)

Answer 2

This is a tricky one.
Best I can come up with is this, which seems to work but I might be missing some edge cases.
SELECT sub0.restaurant_id, MIN(sub1.day)
FROM
(
SELECT restaurant_id, MIN( LEAST(ABS(DAYOFWEEK(CURDATE()) - day), ABS(DAYOFWEEK(CURDATE()) - (day + 7)), ABS(DAYOFWEEK(CURDATE()) - (day - 7)))) AS difference
FROM opening_hours
GROUP BY restaurant_id
) sub0
INNER JOIN
(
SELECT restaurant_id, day, LEAST(ABS(DAYOFWEEK(CURDATE()) - day), ABS(DAYOFWEEK(CURDATE()) - (day + 7)), ABS(DAYOFWEEK(CURDATE()) - (day - 7))) AS difference
FROM opening_hours
) sub1
ON sub0.restaurant_id = sub1.restaurant_id
AND sub0.difference = sub1.difference
GROUP BY sub0.restaurant_id
The first sub query is getting the absolute difference between todays day and each restaurant day. It is using the day, the day plus 7 and the day minus 7 to compare with, using ABS to just get the difference in days and using LEAST to get the lowest of those differences. This way if the current day is 1 and there is a restaurant day of 6 it is comparing 1 + 7 with 6, 1 - 7 with 6 and 1 with 6 and getting the least of those (in this case that would be 1 + 7).
The 2nd sub query just gets that difference and the day of the week for each possible restaurant / day, and this is joined to the first sub query.
The outer query uses MIN just to pick a single day when 2 are just as close.

Answer 3

Something like that should do it:
SELECT oh1.*
-- set the start day
FROM (SELECT #start := 1) AS start,
-- calculate difference in days
(SELECT *, (CASE WHEN day-#start >= 0 THEN day-#start ELSE day-#start+7 END) AS diff
FROM opening_hours) AS oh1
-- find minimum difference
JOIN (SELECT restaurant_id, MIN(CASE WHEN day-#start >= 0 THEN day-#start ELSE day-#start+7 END) AS min_diff
FROM opening_hours
GROUP BY restaurant_id) AS oh2
ON oh1.restaurant_id = oh2.restaurant_id AND
oh1.diff = oh2.min_diff
Replace #start := 1 with your starting day or a call to DAYOFWEEK(CURDATE()), depending on how you want to do it!

Answer 4

First step is to simply add 7 to the result of your day difference calculation when the day is less than the day you are searching for:
SET #Day = 6;
SELECT ID, Restaurant_id, Day,
CASE WHEN Day < #Day THEN 7 ELSE 0 END + Day - #Day AS DaysFromNow
FROM opening_hours;
This will give:
ID RESTAURANT_ID DAY DAYSFROMNOW
1 1 1 2
2 1 4 5
3 2 4 5
4 2 5 6
5 3 4 5
6 3 5 6
7 3 1 2
8 3 6 0
Then to get the next relavant day you need to get the minimum DaysFromNow for each restaurant, then join back to your main table:
SET #Day = 6;
SELECT o.*
FROM opening_hours AS o
INNER JOIN
( SELECT Restaurant_id,
MIN(CASE WHEN Day < #Day THEN 7 ELSE 0 END + Day - #Day) AS DaysFromNow
FROM opening_hours
GROUP BY Restaurant_id
) AS mo
ON mo.Restaurant_id = o.Restaurant_id
AND mo.DaysFromNow = (CASE WHEN Day < #Day THEN 7 ELSE 0 END + Day - #Day);
Example on SQL Fiddle