Finding the lowest node in a given sub-tree of hierarchical data - mysql

Question

Consider a database with these three tables:
category:
cat_id name parent_id
-----------------------
1 drinks 0
2 carbonated 1
3 cola 2
4 water 1
5 rc-cola 3
product:
prod_id name default_cat
-----------------------------------
1 cola-zero 2
2 mineral water 4
cat_prod:
cat_id prod_id
---------------
1 1
2 1
3 1
4 2
We have category hierarchy and a product, which may belong to several categories.
Also, each product has a default category. In this case cola-zero product has default category 2 - carbonated, which is a mistake. Default category has to be 3 - cola. I.e., the lowest category in the category tree. However, I may consider only a subset of the category tree: only those categories that the product belongs to.
I need to update the default category of each product in the product table and ensure that product's default category is the most "defined" one, i.e., the lowest for a given product.
I can write a script, which would retrieve all categories, build the tree in memory and then for each product check the default category against this tree. But I hope there is a smarter way to do this via SQL only.
Is it even possible to do it in pure SQL?
Thanks.

Answer 1

If you store the hierarchy in a Closure Table, it's really easy to find the lowest node(s) in the tree:
SELECT c.descendant FROM closure c
JOIN (SELECT MAX(pathlength) AS pathlength FROM closure) x USING (pathlength);
Finding the lowest node of a subtree, you just need to be specific about the starting node of the branch you want to search:
SELECT c.descendant FROM closure c
JOIN (SELECT MAX(pathlength) AS pathlength FROM closure) x USING (pathlength)
WHERE c.ancestor = 2;

Answer 2

I finally solved it. A bit dirty, since I have to create a temporary table to hold intermediate results, but it works.
Here is the full code:
-- schema
CREATE TABLE category
(
cat_id INT NOT NULL,
name VARCHAR(255) NOT NULL,
parent_id INT NOT NULL,
PRIMARY KEY (cat_id)
);
GO
CREATE TABLE product
(
prod_id INT NOT NULL,
name VARCHAR(255) NOT NULL,
default_cat INT NOT NULL,
PRIMARY KEY (prod_id)
);
GO
CREATE TABLE cat_prod
(
cat_id INT NOT NULL,
prod_id INT NOT NULL,
PRIMARY KEY (cat_id, prod_id),
FOREIGN KEY (cat_id) REFERENCES category(cat_id),
FOREIGN KEY (prod_id) REFERENCES product(prod_id)
);
GO
-- data
INSERT INTO category (cat_id, name, parent_id)
VALUES
(1, 'drinks', 0),
(2, 'carbonated', 1),
(3, 'cola', 2),
(4, 'water', 1),
(5, 'rc-cola', 3)
;
GO
INSERT INTO product (prod_id, name, default_cat)
VALUES
(1, 'cola-zero', 2), -- this is a mistake! must be 3
(2, 'mineral water', 4) -- this one should stay intact
;
GO
INSERT INTO cat_prod (cat_id, prod_id)
VALUES
(1, 1),
(2, 1),
(3, 1),
(4, 2),
(4, 1)
;
GO
-- stored proc
CREATE PROCEDURE iterate_products()
BEGIN
DECLARE prod_id INT;
DECLARE default_cat INT;
DECLARE new_default_cat INT;
DECLARE done INT DEFAULT FALSE;
DECLARE cur CURSOR FOR SELECT p.prod_id, p.default_cat FROM product p;
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = TRUE;
-- temporary table to hold the category subtree for a given product
CREATE TABLE IF NOT EXISTS tmp_category_sub_tree
(
cat_id INT NOT NULL,
parent_id INT NOT NULL
);
OPEN cur;
UPDATE_LOOP: LOOP
FETCH cur INTO prod_id, default_cat;
IF done THEN
LEAVE UPDATE_LOOP;
END IF;
TRUNCATE TABLE tmp_category_sub_tree;
-- select all cateries this products belongs to
INSERT INTO tmp_category_sub_tree (cat_id, parent_id)
SELECT category.cat_id, category.parent_id
FROM category
INNER JOIN cat_prod
ON category.cat_id = cat_prod.cat_id
WHERE
cat_prod.prod_id = prod_id;
-- select a leaf (only one)
SELECT t1.cat_id FROM
tmp_category_sub_tree AS t1 LEFT JOIN tmp_category_sub_tree AS t2
ON t1.cat_id = t2.parent_id
WHERE
t2.cat_id IS NULL
LIMIT 1
INTO NEW_DEFAULT_CAT;
-- update product record, if required
IF default_cat != new_default_cat THEN
UPDATE product
SET default_cat = new_default_cat
WHERE
product.prod_id = prod_id;
END IF;
END LOOP;
CLOSE cur;
DROP TABLE tmp_category_sub_tree;
END;
GO
Here is the SQLFiddle link: http://sqlfiddle.com/#!2/98a45/1