如何在SQLAlchemy中使用collection_class?

spb*_*bks 6 python sqlalchemy

我正在尝试对 SQLAlchemy 中机构参与者之间的层次结构和历史关系进行建模(即机构可以有父母/孩子和前任/继任者)。到目前为止,我所掌握的内容很大程度上遵循SQLAlchemy 文档中的有向图示例。现在我希望能够访问字典中节点的左/右邻居作为edge_type键和节点列表作为值,如下所示:node.right_nodes['edge_type']。

我认为这可以通过 collection_class 来完成,但collection_class=attribute_mapped_collection('edge_type')仅使用单个键:值对的结果,而不是键:[值列表]。

实际结果:

>>> node.right_edges['edge_type']
<Edge object>
>>> node.right_nodes['edge_type']
<Node object>
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预期结果:

>>> node.right_edges['edge_type']
[<Edge object>, <Edge object>]
>>> node.right_nodes['edge_type']
[<Node object>, <Node object>]
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该模型如下所示:

from sqlalchemy import (Column, Integer, String, ForeignKey,
                        create_engine)
from sqlalchemy.orm import Session, relationship, backref
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.ext.associationproxy import association_proxy
from sqlalchemy.orm.collections import attribute_mapped_collection


Base = declarative_base()


class Node(Base):
    __tablename__ = 'node'

    id = Column(Integer, primary_key=True)
    name = Column(String, nullable=False)

    left_nodes = association_proxy('left_edges', 'left_node')
    right_nodes = association_proxy('right_edges', 'right_node')


class Edge(Base):
    __tablename__ = 'edge'

    left_node_id = Column(Integer, ForeignKey('node.id'), primary_key=True)
    right_node_id = Column(Integer, ForeignKey('node.id'), primary_key=True)
    edge_type = Column(String)

    left_node = relationship(
        Node,
        foreign_keys=left_node_id,
        backref=backref(
            'right_edges',
            collection_class=attribute_mapped_collection('edge_type')
        )
    )

    right_node = relationship(
        Node,
        foreign_keys=right_node_id,
        backref=backref(
            'left_edges',
            collection_class=attribute_mapped_collection('edge_type')
        )
    )
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像这样使用:

engine = create_engine('sqlite://', echo=True)
Base.metadata.create_all(engine)
session = Session(engine)

n1 = Node(name='LeftNode')
n2 = Node(name='RightNode1')
n3 = Node(name='RightNode2')
Edge(left_node=n1, right_node=n2, edge_type='hierarchy')
Edge(left_node=n1, right_node=n3, edge_type='hierarchy')
session.add_all([n1, n2, n3])
session.commit()

print(n1.right_nodes)  # returns dict with 1 node as value
print(n1.right_nodes['hierarchy'])  # returns 1 node
print(n1.right_edges)  # returns dict with 1 edge as value
print(n1.right_edges['hierarchy'])  # returns 1 edge
print(session.query(Edge).filter_by(left_node=n1).all())  # returns list of 2 edges
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编辑:

以下内容不是我问题的答案,而是记录我迄今为止所掌握的内容。

association_proxy不仅以字段为目标,还可以在目标类中定义属性:

class Node(Base):

    # <snip>

    left_nodes = association_proxy('left_edges', 'left_nodes_edge_type')
    right_nodes = association_proxy('right_edges', 'right_nodes_edge_type')


class Edge(Base):

    # <snip>

    @property
    def left_nodes_edge_type(self):
        return {self.edge_type: self.left_node}

    @property
    def right_nodes_edge_type(self):
        return {self.edge_type: self.left_node}
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这当然不会产生所需的列表字典,而是一个字典列表:

>>> node.right_nodes
[{'edge_type': <Node object>}, {'edge_type': <Node object>}]
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您还可以简单地在类本身上定义属性,而根本Node不使用它来代理 from to :association_proxyNodeNode

class Node(Base):

    # <snip>

    @property
    def left_nodes(self):
        d = defaultdict(list)
        for edge in self.left_edges:
            d[edge.edge_type].append(edge.left_node)
        return d

    @property
    def right_nodes(self):
        d = defaultdict(list)
        for edge in self.right_edges:
            d[edge.edge_type].append(edge.right_node)
        return d
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这允许所需的视图作为列表的字典:

>>> node.right_nodes
{'edge_type': [<Node object>, <Node object>]}
>>> node.right_nodes['edge_type']
[<Node object>, <Node object>]
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但这只是一种方便的看法。