通过递归CTE获取具有SQLAlChemy的层次结构的顶级父ID [英] Get top level parent id of a hierarchy with SQLAlchemy via recursive CTE

问题描述

我有这样的案例：

|               Note table               |
|---------------------|------------------|
|          id         |     parent_id    |
|---------------------|------------------|
|          1          |     Null         |
|---------------------|------------------|
|          2          |      1
|---------------------|------------------|
|          3          |      2
|---------------------|------------------|
|          4          |      3
|---------------------|------------------|

我想要实现的是获取顶级父ID。 在本例中，如果我传递ID号4，我将获得ID 1，因为ID 1是顶级父级。 当它达到parent_id上的空值时，表示该id是顶级父级。

我试过了，但返回的是我传递给函数的ID。

  def get_top_level_Note(self, id: int):


        hierarchy = self.db.session.query(Note).filter(Note.id == id).cte(name="hierarchy", recursive=True)

        parent = aliased(hierarchy, name="p")
        children = aliased(Note, name="c")

        hierarchy = hierarchy.union_all(self.db.session.query(children).filter(children.parent_id == parent.c.id))

        result = self.db.session.query(Note).select_entity_from(hierarchy).all()

推荐答案

具有名为&Quot；Note&Quot；的现有表

id          parent_id
----------- -----------
11          NULL
22          11
33          22
44          33
55          NULL
66          55

在PostgreSQL中稍作改动就会发现

WITH RECURSIVE parent (i, id, parent_id)
AS (
    SELECT 0, id, parent_id FROM note WHERE id=44
UNION ALL
    SELECT i + 1, n.id, n.parent_id 
    FROM note n INNER JOIN parent p ON p.parent_id = n.id 
    WHERE p.parent_id IS NOT NULL
)
SELECT * FROM parent ORDER BY i;

返回

i           id          parent_id
----------- ----------- -----------
0           44          33
1           33          22
2           22          11
3           11          NULL

，因此我们可以通过将最后一行更改为

来获得顶级父级

WITH RECURSIVE parent (i, id, parent_id)
AS (
    SELECT 0, id, parent_id FROM note WHERE id=44
UNION ALL
    SELECT i + 1, n.id, n.parent_id 
    FROM note n INNER JOIN parent p ON p.parent_id = n.id 
    WHERE p.parent_id IS NOT NULL
)
SELECT id FROM parent ORDER BY i DESC LIMIT 1 ;

返回

id
-----------
11

因此，要将其转换为SQLAlChemy(1.4)：

from sqlalchemy import (
    create_engine,
    Column,
    Integer,
    select,
    literal_column,
)
from sqlalchemy.orm import declarative_base

connection_uri = "postgresql://scott:tiger@192.168.0.199/test"
engine = create_engine(connection_uri, echo=False)

Base = declarative_base()


class Note(Base):
    __tablename__ = "note"
    id = Column(Integer, primary_key=True)
    parent_id = Column(Integer)


def get_top_level_note_id(start_id):
    note_tbl = Note.__table__
    parent_cte = (
        select(
            literal_column("0").label("i"), note_tbl.c.id, note_tbl.c.parent_id
        )
        .where(note_tbl.c.id == start_id)
        .cte(name="parent_cte", recursive=True)
    )
    parent_cte_alias = parent_cte.alias("parent_cte_alias")
    note_tbl_alias = note_tbl.alias()
    parent_cte = parent_cte.union_all(
        select(
            literal_column("parent_cte_alias.i + 1"),
            note_tbl_alias.c.id,
            note_tbl_alias.c.parent_id,
        )
        .where(note_tbl_alias.c.id == parent_cte_alias.c.parent_id)
        .where(parent_cte_alias.c.parent_id.is_not(None))
    )
    stmt = select(parent_cte.c.id).order_by(parent_cte.c.i.desc()).limit(1)
    with engine.begin() as conn:
        result = conn.execute(stmt).scalar()
    return result


if __name__ == "__main__":
    test_id = 44
    print(
        f"top level id for note {test_id} is {get_top_level_note_id(test_id)}"
    )
    # top level id for note 44 is 11

    test_id = 66
    print(
        f"top level id for note {test_id} is {get_top_level_note_id(test_id)}"
    )
    # top level id for note 66 is 55

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