sql – 递归查询挑战 – 简单的父/子示例
发布时间:2020-05-25 07:15:27 所属栏目:MsSql 来源:互联网
导读:注意:在# postgresql上的RhodiumToad的帮助下,我已经找到了一个解决方案,我将其作为答案发布.如果有人可以改善这个,请进来! 我没有能够将previous recursive query solution修改为包含多个“根”(无祖先)节点的以下有向非循环图.我正在尝试编写一个查询,其
|
注意:在# postgresql上的RhodiumToad的帮助下,我已经找到了一个解决方案,我将其作为答案发布.如果有人可以改善这个,请进来! 我没有能够将previous recursive query solution修改为包含多个“根”(无祖先)节点的以下有向非循环图.我正在尝试编写一个查询,其输出通常被称为闭包表:一个多对多表,将每个节点的每个路径存储到每个后代及其自身: 1 2 11 8 4 5 7
/ | | | /
3 | 6
| /
9 | 10
/ /
12 13
/
14
CREATE TABLE node (
id SERIAL PRIMARY KEY,node_name VARCHAR(50) NOT NULL
);
CREATE TABLE node_relations (
id SERIAL PRIMARY KEY,ancestor_node_id INT REFERENCES node(id),descendant_node_id INT REFERENCES node(id)
);
INSERT into node (node_name)
SELECT 'node ' || g FROM generate_series(1,14) g;
INSERT INTO node_relations(ancestor_node_id,descendant_node_id) VALUES
(1,3),(2,(4,6),(5,(7,(3,9),(6,10),(8,(9,12),(11,(10,13),(12,14),(13,14);
很难确定问题 – 我是否缺少node_relation行?查询错了吗? WITH RECURSIVE node_graph AS ( SELECT ancestor_node_id,ARRAY[descendant_node_id] AS path,0 AS level FROM node_relations UNION ALL SELECT nr.ancestor_node_id,ng.path || nr.descendant_node_id,ng.level + 1 AS level FROM node_graph ng JOIN node_relations nr ON nr.descendant_node_id = ng.ancestor_node_id ) SELECT path[array_upper(path,1)] AS ancestor,path[1] AS descendant,path,level as depth FROM node_graph ORDER BY level,ancestor; 预期产量: ancestor | descendant | path
---------+------------+------------------
1 | 3 | "{1,3}"
1 | 9 | "{1,3,9}"
1 | 12 | "{1,9,12}"
1 | 14 | "{1,12,14}"
2 | 3 | "{2,3}"
2 | 9 | "{2,9}"
2 | 12 | "{2,12}"
2 | 14 | "{2,14}"
3 | 9 | "{3,9}"
3 | 12 | "{3,12}"
3 | 14 | "{3,14}"
4 | 6 | "{4,6}"
4 | 10 | "{4,6,10}"
4 | 13 | "{4,10,13}"
4 | 14 | "{4,13,14}"
5 | 6 | "{5,6}"
5 | 10 | "{5,10}"
5 | 13 | "{5,13}"
5 | 14 | "{5,14}"
6 | 10 | "{6,10}"
6 | 13 | "{6,13}"
6 | 14 | "{6,14}"
7 | 6 | "{7,6}"
7 | 10 | "{7,10}"
7 | 13 | "{7,13}"
7 | 14 | "{7,14}"
8 | 10 | "{8,10}"
8 | 13 | "{8,13}"
8 | 14 | "{8,14}"
9 | 12 | "{9,12}"
9 | 14 | "{9,14}"
10 | 13 | "{10,13}"
10 | 14 | "{10,14}"
11 | 12 | "{11,12}"
11 | 14 | "{11,14}"
12 | 14 | "{12,14}"
13 | 14 | "{13,14}"
解决方法在#postgresql的RhodiumToad帮助下,我已经到达了这个解决方案:WITH RECURSIVE node_graph AS (
SELECT ancestor_node_id as path_start,descendant_node_id as path_end,array[ancestor_node_id,descendant_node_id] as path
FROM node_relations
UNION ALL
SELECT ng.path_start,nr.descendant_node_id as path_end,ng.path || nr.descendant_node_id as path
FROM node_graph ng
JOIN node_relations nr ON ng.path_end = nr.ancestor_node_id
)
SELECT * from node_graph order by path_start,array_length(path,1);
结果与预期完全一样. (编辑:安卓应用网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
