Per group, find first N users with SUM(x) >= N - mysql

Question

Problem: Find the first 2 users who have at least 10 items in a category, per category.
Table structure:
CREATE TABLE items(
id INT AUTO_INCREMENT PRIMARY KEY,
datetime datetime,
category INT,
user INT,
items_count INT
);
Sample data:
INSERT INTO items (datetime, category, user, items_count) VALUES
('2013-01-01 00:00:00', 1, 1, 10),
('2013-01-01 00:00:01', 1, 2, 1),
('2013-01-01 00:00:02', 1, 3, 10),
('2013-01-01 00:00:03', 1, 2, 9),
('2013-01-01 00:00:00', 2, 4, 10),
('2013-01-01 00:00:01', 2, 1, 10),
('2013-01-01 00:00:01', 2, 5, 10);
Desired result:
category user
1 1
1 3
2 4
2 5
Note: As shown in the result, I need to be able to show preference towards a user when multiple users meet the requirements simultaneously.
SQL Fiddle:
http://sqlfiddle.com/#!2/58e60
This is what I have tried:
SELECT
Derived.*,
IF (#category != Derived.category, #rank := 1, #rank := #rank + 1) AS rank,
#category := category
FROM(
SELECT
category,
user,
SUM(items_count) AS items_count,
MAX(datetime) AS datetime
FROM items
GROUP BY
category,
user
HAVING
SUM(items_count) >= 10
) AS Derived
JOIN(SELECT #rank := 0, #category := 0) AS r
HAVING
rank <= 2
ORDER BY
Derived.category,
Derived.datetime
But it is faulty. Not only does it not take user precedence into account, it would produce the wrong result with data such as this:
('2013-01-01 00:00:00', 1, 1, 10),
('2013-01-01 00:00:01', 1, 2, 1),
('2013-01-01 00:00:02', 1, 3, 10),
('2013-01-01 00:00:03', 1, 2, 9),
('2013-01-01 00:00:10', 1, 3, 1);
Additional information: I do not know if procedures could make a difference in this scenario, but unfortunately it is not an option either. The user running this query only has SELECT privilege.

Answer 1

In order to find the users that meet your needs, you need the cumulative sum of the counts. The following query finds the occasions when a user first reaches 10 units. If the counts are never negative, then there is only one:
select i.*
from (select i.*,
(select sum(items_count)
from items i2
where i2.user = i.user and
i2.category = i.category and
i2.datetime <= i.datetime
) as cumsum
from items i
) i
where cumsum - items_count < 10 and cumsum >= 10
order by datetime;
To get the first two, you need to use MySQL tricks for counting within a group. Here is an example that generally works:
select i.*
from (select i.*, if(#prevc = category, #rn := #rn + 1, #rn := 1) as rn, #prevc := category
from (select i.*,
(select sum(items_count)
from items i2
where i2.user = i.user and
i2.category = i.category and
i2.datetime <= i.datetime
) as cumsum
from items i
) i
cross join
(select #rn := 0) const
where cumsum - items_count < 10 and cumsum >= 10
) i
where rn <= 2
order by category, datetime;
I have a problem with this approach, because nothing in MySQL says that the expression #prevc := category will actually be calculated after the calculation for rn. However, it seems to be the case, and this seems to work in practice.

Answer 2

I tried Gordon's query, but unfortunately it does not seem to work with large tables; after waiting 15 minutes for the result I decided to kill it.
However the following query worked very well for me, it chewed it's way through a table of ~6M rows in about 8 seconds.
#Variable
SET #min_items = 10,
#max_users = 2,
#preferred_user = 5,
#Static
#category = 0,
#user = 0,
#items = 0,
#row_num = 1;
--
SELECT
category,
user,
datetime
FROM(
SELECT
category,
user,
datetime,
IF (#category = category, #row_num := #row_num + 1, #row_num := 1) AS row_num,
#category := category
FROM(
SELECT
category,
user,
datetime,
IF (#user != user, #items := 0, NULL),
IF (#items < #min_items, #items := #items + items_count, NULL) AS items_cumulative,
#user := user
FROM items
ORDER BY
category,
user,
datetime
) AS Derived
WHERE items_cumulative >= #min_items
ORDER BY
category,
datetime,
FIELD(user, #preferred_user, user)
) AS Derived
WHERE row_num <= #max_users;