Sql 公共表表达式内部的反向聚合

我希望下面的查询返回所有人及其各自的孩子

WITH RECURSIVE nested_people (id, name, children) AS (
   SELECT id, name, NULL::JSON AS children
   FROM people
   WHERE parent_id IS NULL
 UNION ALL
   SELECT people.id, people.name, ROW_TO_JSON(nested_people.*) AS children
   FROM people
   JOIN nested_people ON people.parent_id = nested_people.id
)
SELECT * FROM nested_people;

但实际上情况正好相反。我想不出一种不需要额外的CTE就能正确嵌套的方法。有办法吗

示例数据

+----+-------+-----------+
| id | name  | parent_id |
+----+-------+-----------+
|  1 | Adam  | null      |
|  2 | Abel  | 1         |
|  3 | Cain  | 1         |
|  4 | Enoch | 3         |
+----+-------+-----------+

结果

+----+-------+--------------------------------------------------------------------------+
| id | name  |                                children                                  |
+----+-------+--------------------------------------------------------------------------+
|  1 | Adam  | null                                                                     |
|  2 | Abel  | {"id":1,"name":"Adam","children":null}                                   |
|  3 | Cain  | {"id":1,"name":"Adam","children":null}                                   |
|  4 | Enoch | {"id":3,"name":"Cain","children":{"id":1,"name":"Adam","children":null}} |
+----+-------+--------------------------------------------------------------------------+

+----+-------+----------------------------------------------------------------------------------------------------------------------+
| id | name  |                                                       children                                                       |
+----+-------+----------------------------------------------------------------------------------------------------------------------+
|  1 | Adam  | [{"id":2, "name":"Abel", "children":null},{"id":3,"name":"Cain","children":[{"id":4,"name":"Enoch","children":null}] |
|  2 | Abel  | null                                                                                                                 |
|  3 | Cain  | [{"id":4,"name":"Enoch","children":null}]                                                                            |
|  4 | Enoch | null                                                                                                                 |
+----+-------+----------------------------------------------------------------------------------------------------------------------+

预期结果

+----+-------+--------------------------------------------------------------------------+
| id | name  |                                children                                  |
+----+-------+--------------------------------------------------------------------------+
|  1 | Adam  | null                                                                     |
|  2 | Abel  | {"id":1,"name":"Adam","children":null}                                   |
|  3 | Cain  | {"id":1,"name":"Adam","children":null}                                   |
|  4 | Enoch | {"id":3,"name":"Cain","children":{"id":1,"name":"Adam","children":null}} |
+----+-------+--------------------------------------------------------------------------+

+----+-------+----------------------------------------------------------------------------------------------------------------------+
| id | name  |                                                       children                                                       |
+----+-------+----------------------------------------------------------------------------------------------------------------------+
|  1 | Adam  | [{"id":2, "name":"Abel", "children":null},{"id":3,"name":"Cain","children":[{"id":4,"name":"Enoch","children":null}] |
|  2 | Abel  | null                                                                                                                 |
|  3 | Cain  | [{"id":4,"name":"Enoch","children":null}]                                                                            |
|  4 | Enoch | null                                                                                                                 |
+----+-------+----------------------------------------------------------------------------------------------------------------------+

---

此rCTE从另一侧穿过树：

WITH RECURSIVE cte AS (
   SELECT id, parent_id, name, NULL::JSON AS children
   FROM   people p
   WHERE  NOT EXISTS (  -- only leaf nodes; see link below
      SELECT 1 FROM people
      WHERE  parent_id = p.id
      )
   UNION ALL
   SELECT p.id, p.parent_id, p.name, row_to_json(c) AS children
   FROM   cte c
   JOIN   people p ON p.id = c.parent_id
   )
SELECT id, name, json_agg(children) AS children
FROM   cte
GROUP  BY 1, 2;

用于聚合外部

选择中每个节点的多个分支

与预期结果的细微差异：


这包括
children
列中的
parent\u id
单个节点未包装到数组中

两者都可以调整，但我希望结果对你来说都是可以的

如何识别叶节点：



您能添加一些数据样本吗？从这个样本中可以得到什么样的结果？谢谢，好主意。我补充了你要求的内容。