Question

我有一个表示直接祖先（不是树结构）的pilates_bill表

bill_id (pk) | previous_bill_id (self-join fk)
=============+================================
1               2
2               3
3               4
5               NULL

需要为任何给定行生成所有祖先的列表（父/祖父/ /等）（以下示例以1开头）。

使用递归CTE获取具有祖先链的bill_id列表

WITH RECURSIVE chain(from_id, to_id) AS (
  SELECT NULL::integer, 1  -- starting id
  UNION
  SELECT c.to_id, pilates_bill.previous_bill_id
  FROM chain c
  LEFT OUTER JOIN pilates_bill ON (pilates_bill.bill_id = to_id)
  WHERE c.to_id IS NOT NULL
)
SELECT from_id FROM chain WHERE from_id IS NOT NULL;

结果1,2,3,4,5符合预期

但是现在当我尝试按祖先的顺序生成表行时，结果会被破坏

SELECT * FROM pilates_bill WHERE bill_id IN
(
WITH RECURSIVE chain(from_id, to_id) AS (
  SELECT NULL::integer, 1
  UNION
  SELECT c.to_id, pilates_bill.previous_bill_id
  FROM chain c
  LEFT OUTER JOIN pilates_bill ON (pilates_bill.bill_id = to_id)
  WHERE c.to_id IS NOT NULL
)
SELECT from_id FROM chain WHERE from_id IS NOT NULL
)

行顺序为5,1,2,3,4

我在这做错了什么？

Answer 1

SQL查询返回的行是随机顺序，除非您指定order by。

您可以通过在递归CTE中跟踪深度来计算深度：

WITH RECURSIVE chain(from_id, to_id, depth) AS
    (
    SELECT  NULL::integer
    ,       1
    ,       1
    UNION
    SELECT  c.to_id
    ,       pb.previous_bill_id
    ,       depth + 1
    FROM chain c
    LEFT JOIN 
            pilates_bill pb
    ON      pb.bill_id = c.to_id
    WHERE   c.to_id IS NOT NULL
    )
SELECT  * 
FROM    chain
ORDER BY
        depth

Postgres自连接递归CTE祖先链

1 个答案: