递归内加入与邻接

Question

我有这样的事情：

CREATE TABLE categories (
id varchar(250) PRIMARY KEY, 
name varchar(250) NOT NULL, 
parentid varchar(250) 
); 

CREATE TABLE products (
id varchar(250) PRIMARY KEY, 
name varchar(250) NOT NULL, 
price double precision, 
category varchar(250) NOT NULL 
); 

INSERT INTO categories VALUES ('1', 'Rack', ''); 
INSERT INTO categories VALUES ('2', 'Women', '1'); 
INSERT INTO categories VALUES ('3', 'Shorts', '2'); 

INSERT INTO products VALUES ('1', 'Jean', 2.99, '3'); 
INSERT INTO products VALUES ('2', 'Inflatable Boat', 5.99, '1'); 

现在，如果我想看到的产品的总价格为每个类别，我可以做这样的事情：

SELECT 
categories.name, 
SUM(products.price) AS CATPRICE 
FROM 
categories, 
products 
WHERE products.category = categories.id 
GROUP BY categories.name 
; 

将会产生输出：

name | catprice 
--------+---------- 
Rack |  5.99 
Shorts |  2.99 
(2 rows) 

但是请注意，“短裤”是“机架”的祖先。我想要一个查询，将产生这样的输出：

name | catprice 
--------+---------- 
Rack |  8.98 
(1 row) 

让所有的产品价格是根类别下加在一起。类别表中有多个根类别;为简单起见，只有一个。

这是我迄今：

-- "nodes_cte" is the virtual table that is being created as the recursion continues 
-- The contents of the()s are the columns that are being built 
WITH RECURSIVE nodes_cte(name, id, parentid, depth, path) AS (
-- Base case? 
SELECT tn.name, tn.id, tn.parentid, 1::INT AS depth, tn.id::TEXT AS path FROM categories AS tn, products AS tn2 
LEFT OUTER JOIN categories ON tn2.CATEGORY = categories.ID 
WHERE tn.parentid IS NULL 
UNION ALL 
-- nth case 
SELECT c.name, c.id, c.parentid, p.depth + 1 AS depth, (p.path || '->' || c.id::TEXT) FROM nodes_cte AS p, categories AS c, products AS c2 
LEFT OUTER JOIN categories ON c2.CATEGORY = categories.ID 
WHERE c.parentid = p.id 
) 
SELECT * FROM nodes_cte AS n ORDER BY n.id ASC; 

我不知道我做了什么错。上面的查询返回零结果。

Answer 1

您的递归查询已关闭。这给一试：

编辑 - 为了与SUM这项工作，使用：

WITH RECURSIVE nodes_cte(name, id, id2, parentid, price) AS (
-- Base case? 
SELECT c.name, 
    c.id, 
    c.id id2, 
    c.parentid, 
    p.price 
FROM categories c 
    LEFT JOIN products p on c.id = p.category 
WHERE c.parentid = '' 
UNION ALL 
-- nth case 
SELECT n.name, 
    n.id, 
    c.id id2, 
    c.parentid, 
    p.price 
FROM nodes_cte n 
    JOIN categories c on n.id2 = c.parentid 
    LEFT JOIN products p on c.id = p.category 
) 
SELECT id, name, SUM(price) FROM nodes_cte GROUP BY id, name 

这里是小提琴：http://sqlfiddle.com/#!1/7ac6d/19

好运。