Question

我原以为以下查询会返回所有拥有各自子女的人。

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;

但实际上它完全相反。我无法想出一种无需额外CTE即可正确嵌套的方法。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

此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 Fiddle.

使用json_agg()汇总外部SELECT中每个节点的多个分支与您期望的结果的微小差异：

这包括parent_id列中的children。
单个节点未包装到数组中。

要么可以调整，但我希望结果对你来说没问题。

如何识别叶节点：

Select rows which are not present in other table

公共表表达式内的反向聚合

1 个答案: