0

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，如果我想查看每个类别的产品总价，我可以这样做：

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

产生输出：

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

但请注意，“Shorts”是“Rack”的祖先。我想要一个会产生如下输出的查询：

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

以便将所有产品价格添加到根类别下。类别表中有多个根类别；为简单起见，只显示了一个。

这是我迄今为止所拥有的：

-- "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;

我不知道我做错了什么。上述查询返回零个结果。

4

1 回答 1

0

您的递归查询稍微偏离了一点。试试这个：

编辑——要使用 SUM 进行这项工作，请使用以下命令：

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

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

祝你好运。

于 2013-02-10T22:07:56.673 回答