[英]Postgres self-join recursive CTE ancestry chain
我有一个pilates_bill表代表直接祖先(不是树结构)
bill_id (pk) | previous_bill_id (self-join fk)
=============+================================
1 2
2 3
3 4
5 NULL
需要为任何给定的行生成所有祖先的列表(父母/祖父母/祖父母等)(以下示例以1开头)。
使用递归CTE获取具有祖先链的bill_ids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如预期
但是现在当我尝试按祖先顺序生成表行时,结果就坏了
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
我在这里做错了什么?
除非您指定的order by
否则SQL查询返回的行的顺序是随机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
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