使用查询以递归方式查找父级
使用查询以递归方式查找父级
提问于 2010-09-13 18:03:31
我使用的是postgresql。我的表格如下所示
parent_id child_id
----------------------
101 102
103 104
104 105
105 106 我想写一个sql查询，它将给出输入的最终父元素。
也就是说，假设我将106作为输入，那么它的输出将是103。
(106 --> 105 --> 104 --> 103)回答 2
Stack Overflow用户
发布于 2011-06-22 04:45:46
下面是一个完整的示例。首先是DDL
test=> CREATE TABLE node (
test(> id SERIAL,
test(> label TEXT NOT NULL, -- name of the node
test(> parent_id INT,
test(> PRIMARY KEY(id)
test(> );
NOTICE: CREATE TABLE will create implicit sequence "node_id_seq" for serial column "node.id"
NOTICE: CREATE TABLE / PRIMARY KEY will create implicit index "node_pkey" for table "node"
CREATE TABLE...and一些数据...
test=> INSERT INTO node (label, parent_id) VALUES ('n1',NULL),('n2',1),('n3',2),('n4',3);
INSERT 0 4
test=> INSERT INTO node (label) VALUES ('garbage1'),('garbage2'), ('garbage3');
INSERT 0 3
test=> INSERT INTO node (label,parent_id) VALUES ('garbage4',6);
INSERT 0 1
test=> SELECT * FROM node;
id | label | parent_id
----+----------+-----------
1 | n1 |
2 | n2 | 1
3 | n3 | 2
4 | n4 | 3
5 | garbage1 |
6 | garbage2 |
7 | garbage3 |
8 | garbage4 | 6
(8 rows)这将对node中的每个id执行递归查询：
test=> WITH RECURSIVE nodes_cte(id, label, parent_id, depth, path) AS (
SELECT tn.id, tn.label, tn.parent_id, 1::INT AS depth, tn.id::TEXT AS path
FROM node AS tn
WHERE tn.parent_id IS NULL
UNION ALL
SELECT c.id, c.label, c.parent_id, p.depth + 1 AS depth,
(p.path || '->' || c.id::TEXT)
FROM nodes_cte AS p, node AS c
WHERE c.parent_id = p.id
)
SELECT * FROM nodes_cte AS n ORDER BY n.id ASC;
id | label | parent_id | depth | path
----+----------+-----------+-------+------------
1 | n1 | | 1 | 1
2 | n2 | 1 | 2 | 1->2
3 | n3 | 2 | 3 | 1->2->3
4 | n4 | 3 | 4 | 1->2->3->4
5 | garbage1 | | 1 | 5
6 | garbage2 | | 1 | 6
7 | garbage3 | | 1 | 7
8 | garbage4 | 6 | 2 | 6->8
(8 rows)这将获取node.id =1的所有后代：
test=> WITH RECURSIVE nodes_cte(id, label, parent_id, depth, path) AS (
SELECT tn.id, tn.label, tn.parent_id, 1::INT AS depth, tn.id::TEXT AS path FROM node AS tn WHERE tn.id = 1
UNION ALL
SELECT c.id, c.label, c.parent_id, p.depth + 1 AS depth, (p.path || '->' || c.id::TEXT) FROM nodes_cte AS p, node AS c WHERE c.parent_id = p.id
)
SELECT * FROM nodes_cte AS n;
id | label | parent_id | depth | path
----+-------+-----------+-------+------------
1 | n1 | | 1 | 1
2 | n2 | 1 | 2 | 1->2
3 | n3 | 2 | 3 | 1->2->3
4 | n4 | 3 | 4 | 1->2->3->4
(4 rows)下面将获取id为4的节点的路径：
test=> WITH RECURSIVE nodes_cte(id, label, parent_id, depth, path) AS (
SELECT tn.id, tn.label, tn.parent_id, 1::INT AS depth, tn.id::TEXT AS path
FROM node AS tn
WHERE tn.parent_id IS NULL
UNION ALL
SELECT c.id, c.label, c.parent_id, p.depth + 1 AS depth,
(p.path || '->' || c.id::TEXT)
FROM nodes_cte AS p, node AS c
WHERE c.parent_id = p.id
)
SELECT * FROM nodes_cte AS n WHERE n.id = 4;
id | label | parent_id | depth | path
----+-------+-----------+-------+------------
4 | n4 | 3 | 4 | 1->2->3->4
(1 row)让我们假设您想要将搜索限制在depth小于3的子代(注意，depth尚未递增)：
test=> WITH RECURSIVE nodes_cte(id, label, parent_id, depth, path) AS (
SELECT tn.id, tn.label, tn.parent_id, 1::INT AS depth, tn.id::TEXT AS path
FROM node AS tn WHERE tn.id = 1
UNION ALL
SELECT c.id, c.label, c.parent_id, p.depth + 1 AS depth,
(p.path || '->' || c.id::TEXT)
FROM nodes_cte AS p, node AS c
WHERE c.parent_id = p.id AND p.depth < 2
)
SELECT * FROM nodes_cte AS n;
id | label | parent_id | depth | path
----+-------+-----------+-------+------
1 | n1 | | 1 | 1
2 | n2 | 1 | 2 | 1->2
(2 rows)我建议使用ARRAY数据类型而不是字符串来演示“路径”，但箭头更能说明parent<=>child关系。
Stack Overflow用户
发布于 2017-02-25 04:34:42
使用WITH RECURSIVE创建公用表表达式。对于非递归项，获取子项紧接父项下面的行：
SELECT
c.child_id,
c.parent_id
FROM
mytable c
LEFT JOIN
mytable p ON c.parent_id = p.child_id
WHERE
p.child_id IS NULL
child_id | parent_id
----------+-----------
102 | 101
104 | 103对于递归项，您需要这些子项的子项。
WITH RECURSIVE tree(child, root) AS (
SELECT
c.child_id,
c.parent_id
FROM
mytable c
LEFT JOIN
mytable p ON c.parent_id = p.child_id
WHERE
p.child_id IS NULL
UNION
SELECT
child_id,
root
FROM
tree
INNER JOIN
mytable on tree.child = mytable.parent_id
)
SELECT * FROM tree;
child | root
-------+------
102 | 101
104 | 103
105 | 103
106 | 103您可以在查询CTE时对下级进行过滤：
WITH RECURSIVE tree(child, root) AS (...) SELECT root FROM tree WHERE child = 106;
root
------
103页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/3699395
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