每行的递归总和答案

【问题标题】：Recursive sum for each row每行的递归总和
【发布时间】：2023-02-12 02:34:35
【问题描述】：

考虑这个测试数据：

CREATE TABLE IF NOT EXISTS area (
    id integer,
    parent_id integer,
    name text,
    population integer
);

INSERT INTO area VALUES
    (1, NULL, 'North America', 0),
    (2, 1, 'United States', 0),
    (3, 1, 'Canada', 39),
    (4, 1, 'Mexico', 129),
    (5, 2, 'Contiguous States', 331),
    (6, 2, 'Non-contiguous States', 2);

id	parent_id	name	population
1	NULL	North America	0
2	1	United States	0
3	1	Canada	39
4	1	Mexico	129
5	2	Contiguous States	331
6	2	Non-contiguous States	2

请注意，population（以百万为单位）在这里表示增加的人口，不包括该地区的儿童。

如何查询每一行的递归和？我需要得到这样的东西：

name	sum
North America	501
United States	333
Canada	39
Mexico	129
Contiguous States	331
Non-contiguous States	2

【问题讨论】：

标签： postgresql common-table-expression recursive-query

【解决方案1】：

这是我解决它的方法：

WITH RECURSIVE parent AS (
    SELECT
        *,
        id AS root  -- Emit the root ID
    FROM area
    -- No WHERE clause, since we treat every row as a possible root
    
    UNION ALL   
    
    SELECT
        area.*,
        parent.root  -- Pass the root ID to the children
    FROM area
    JOIN parent ON parent.id = area.parent_id
)
SELECT
    name,
    sum
FROM (
    SELECT
        root,
        SUM(population)
    FROM parent
    GROUP BY root  -- Here is where we need the root IDs
) cumulative
JOIN area ON area.id = cumulative.root;  -- Use the nice names instead of IDs

【讨论】：