Thank you very much .
aliased (TreeNode, row_expression) was exactly what I was looking for.
Never noticed that before.
On Tuesday, November 10, 2020 at 12:51:30 PM UTC-8 Mike Bayer wrote:
> I would need to see the table defs for "docorder" to do this completely.
>
> The parents + kids version:
>
> # 1. make parents cte
>
> parents_cte = session.query(TreeNode).filter(TreeNode.name ==
> "subnode4").cte("parents", recursive=True)
> p2 = session.query(TreeNode).filter(TreeNode.id == parents_cte.c.parent_id)
> parents_cte = parents_cte.union(p2)
>
> # 2. make kids cte
>
> kids_cte = session.query(TreeNode).filter(TreeNode.name ==
> "subnode4").cte("kids", recursive=True)
> k2 = session.query(TreeNode).filter(TreeNode.parent_id ==
> kids_cte.c.parent_id)
> kids_cte = kids_cte.union(k2)
>
> # 3. SELECT from both CTEs and create a union, then a subquery
> from sqlalchemy import union
> treenode_rows = union(parents_cte.select(), kids_cte.select()).alias("n")
>
> 4. SELECT TreeNode entities from the subuqery (docs:
> https://docs.sqlalchemy.org/en/13/orm/tutorial.html#selecting-entities-from-subqueries)
>
>
>
> from sqlalchemy.orm import aliased
> treenode_alias = aliased(TreeNode, treenode_rows)
>
> # 5. select rows
> print(session.query(treenode_alias).order_by(treenode_alias.id).all())
>
> this looks like exactly your first query if I'm not mistaken and it
> returns TreeNode objects per aliased(). hope this gets you started.
>
>
>
>
> On Tue, Nov 10, 2020, at 2:55 PM, kris wrote:
>
>
> A more complete version of the SQL to be returned as TreeNode
>
> WITH RECURSIVE
> docorder AS ( select id, rn from ...),
> parents AS
> (SELECT tree.id AS id, tree.parent_id AS parent_id, tree.name AS name
> FROM tree
> WHERE tree.name = 'subnode4'
> UNION SELECT tree.id AS tree_id, tree.parent_id AS tree_parent_id,
> tree.name AS tree_name
> FROM tree, parents
> WHERE tree.id = parents.parent_id),
> kids AS
> (SELECT tree.id AS id, tree.parent_id AS parent_id, tree.name AS name
> FROM tree
> WHERE tree.name = 'subnode4'
> UNION SELECT tree.id AS tree_id, tree.parent_id AS tree_parent_id,
> tree.name AS tree_name
> FROM tree, kids
> WHERE tree.parent_id = kids.id)
> SELECT n.id, n.parent_id, n.name
> FROM (
> SELECT parents.id, parents.parent_id, parents.name
> FROM parents
> UNION
> SELECT kids.id, kids.parent_id, kids.name
> FROM kids) AS n,
> docorder
> WHERE n.id = docorder.id
> ORDER BY docorder.rn, n.parent_id
>
>
>
>
> --
> SQLAlchemy -
> The Python SQL Toolkit and Object Relational Mapper
>
> http://www.sqlalchemy.org/
>
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http://www.sqlalchemy.org/
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