公用表表达式内的反向聚合

Question

我本来希望以下查询返回所有带有各自子代的人。

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;

但是实际上，它的确是相反的。我想不出无需额外的CTE就能进行正确嵌套的方法。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}} |
+----+-------+--------------------------------------------------------------------------+

预期结果

+----+-------+----------------------------------------------------------------------------------------------------------------------+
| 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                                                                                                                 |
+----+-------+----------------------------------------------------------------------------------------------------------------------+

Answer 1

该rCTE从另一侧遍历树：

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;

SQL提琴。

使用json_agg()到每个节点聚集多个分支外SELECT。
与您期望的结果之间的微小差异：

• 这包括parent_id在children列中。
• 没有将单个节点包装到数组中。

都可以进行调整，但是我希望结果对您来说是可以的。