Find sum of stacked/overlapping date intersections in SQL table - mysql

Question

I have the following table which represents bookings of articles:
+---+------------+----------+-------------+-------------+
|id | article_id | quantity | starts_at | ends_at |
+---+------------+----------+-------------+-------------+
| 1 | 1 | 1 | 2015-03-01 | 2015-03-20 |
| 2 | 1 | 2 | 2015-03-02 | 2015-03-03 |
| 3 | 1 | 3 | 2015-03-04 | 2015-03-15 |
| 4 | 1 | 2 | 2015-03-16 | 2015-03-22 |
| 5 | 1 | 2 | 2015-03-11 | 2015-03-19 |
| 6 | 2 | 2 | 2015-03-06 | 2015-03-22 |
| 7 | 2 | 3 | 2015-03-02 | 2015-03-04 |
+---+------------+----------+-------------+-------------+
From this table I want to extract the following information:
+------------+----------+
| article_id | sum |
+------------+----------+
| 1 | 6 |
| 2 | 3 |
+------------+----------+
Sum represents the max sum of quantity of stacked/overlapping booked articles for the given time ranges. In the first table article with id=1 has its maximum from booking 1, 3 and 5.
Is there any MySQL solution to obtain this information from a table like this?
Thank you very much!
EDIT: The date intersections are crucial. Let's say booking 5 starts at 2015-03-17 the sum for article_id=1 results 5, because booking 3 and 5 are not overlapping anymore. The sql should automatically consider all possible overlapping possibilities.

Answer 1

My answer is going to seem crazy complicated, perhaps; but it isn't, if one accepts that the use of a calendar table is an excellent MySQL idiom for dealing with date range related issues. I've closely adapted calendar table code from Artful Software's calendar table article. Artful Software's query techniques are a wonderful resource for doing complicated things in MySQL. The calendar table gives you a row per individual date that you are working with, which makes many things much easier.
For the whole thing below, you can go to this sqlfiddle for a place to play around with the code. It'll take a while to load.
First, here is your data:
CREATE TABLE articles
(`id` int, `article_id` int, `quantity` int, `starts_at` datetime, `ends_at` datetime);
INSERT INTO articles
(`id`, `article_id`, `quantity`, `starts_at`, `ends_at`)
VALUES
(1, 1, 1, '2015-03-01 00:00:00', '2015-03-20 00:00:00'),
(2, 1, 2, '2015-03-02 00:00:00', '2015-03-03 00:00:00'),
(3, 1, 3, '2015-03-04 00:00:00', '2015-03-15 00:00:00'),
(4, 1, 2, '2015-03-16 00:00:00', '2015-03-22 00:00:00'),
(5, 1, 2, '2015-03-11 00:00:00', '2015-03-19 00:00:00'),
(6, 2, 2, '2015-03-06 00:00:00', '2015-03-22 00:00:00'),
(7, 2, 3, '2015-03-02 00:00:00', '2015-03-04 00:00:00');
Next, here is the creation of the calendar table--I've created somewhat more date rows than needed (going back to start of year, and forward to start of next year). Ideally you just permanently keep a more massive calendar table on hand, covering a span of dates that will handle anything you could ever need. All the stuff below is going to seem quite lengthy and complex. But if you already have a calendar table lying around, the whole next section is not necessary.
CREATE TABLE calendar ( dt datetime primary key );
/* the views below will be joined and rejoined to themselves to
get the effect creating many rows. V ends up with 10 rows. */
CREATE OR REPLACE VIEW v3 as SELECT 1 n UNION ALL SELECT 1 UNION ALL SELECT 1;
CREATE OR REPLACE VIEW v as SELECT 1 n FROM v3 a, v3 b UNION ALL SELECT 1;
/* Going to limit the calendar table to first of year of min date
and first of year after max date */
SELECT #min := makedate(year(min(starts_at)),1) FROM articles;
SELECT #max := makedate(year(min(ends_at))+1,1) FROM articles;
SET #inc = -1;
/* below we work with #min date + #inc days successively, with #inc:=#inc+1
acting like ++variable, so we start with minus 1.
We insert as many individual date rows as we want by self-joining v,
and using some kind of limit via WHERE to keep the calendar table small
for our example. For n occurrences of v below, you get a max
of 10^n rows in the calendar table. We are using v as row-creation
engine. */
INSERT INTO calendar
SELECT #min + interval #inc:=#inc+1 day as dt
FROM v a, v b, v c, v d # , v e , v f
WHERE #inc < datediff(#max,#min);
Now we are ready to find the stackings. Assuming the above (big assumption, I know), this becomes pretty easy. I'm going to do it through a few views for readability.
/* now create a view that will let us easily view the articles
related to indvidual dates when we query.
Not necessary, just makes things easier to read. */
CREATE OR REPLACE VIEW articles_to_dates as
SELECT c.dt, article_id
FROM articles a
INNER JOIN calendar c on c.dt between (SELECT min(starts_at) FROM articles) and (SELECT max(ends_at) FROM articles)
GROUP BY article_id, c.dt;
--SELECT * FROM articles_to_dates --This query would show the view's result
/* next view is the total amount of articles booked per individual date */
CREATE OR REPLACE VIEW booked_quantities_per_day AS
SELECT a2d.dt,a2d.article_id, SUM(a.quantity) as booked_quantity
FROM articles_to_dates a2d
INNER JOIN articles a on a2d.dt between a.starts_at and a.ends_at and a.article_id = a2d.article_id
GROUP BY a2d.dt, a2d.article_id
ORDER by a2d.article_id, a2d.dt
--SELECT * from booked_quantities_per_day --this query would show the view's result
Finally, here are the desired results:
SELECT article_id, max(booked_quantity) max_stacked
FROM booked_quantities_per_day
GROUP BY article_id;
Results:
article_id max_stacked
1 6
2 3

Answer 2

This should work.
Two groups. First to get distinct list of possible 'quantity'; second - summarise them
SELECT article_id, SUM(sub.quantity) FROM
(SELECT article_id, quantity FROM table GROUP BY article_id, quantity) as sub
GROUP BY article_id

Answer 3

select sum(quantity) from ...
group by article_id
where
... select your date range ...