python - 如何在 SQLAlchemy 模型的构造函数中通过关系存储数据？

Question

如何在构造函数中添加有关系的对象？评估构造函数时，id 尚未准备好。在更简单的情况下，可以只提供预先计算的列表。在下面的例子中，我试图说有一个complex_cls_method，在某种程度上它更像是黑匣子。

from sqlalchemy import create_engine, MetaData, Column, Integer, String, ForeignKey
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship
from sqlalchemy.orm import sessionmaker

DB_URL = "mysql://user:password@localhost/exampledb?charset=utf8"

engine = create_engine(DB_URL, encoding='utf-8', convert_unicode=True, pool_recycle=3600, pool_size=10)
session = sessionmaker(autocommit=False, autoflush=False, bind=engine)()

Model = declarative_base()


class User(Model):
    __tablename__ = 'user'
    id = Column(Integer, primary_key=True)
    simple = Column(String(255))
    main_address = Column(String(255))
    addresses = relationship("Address",
                             cascade="all, delete-orphan")

    def __init__(self, addresses, simple):
        self.simple = simple
        self.main_address = addresses[0]
        return  # because the following does not work
        self.addresses = Address.complex_cls_method(
            user_id_=self.id,  # <-- this does not work of course
            key_="address",
            value_=addresses
        )


class Address(Model):
    __tablename__ = 'address'
    id = Column(Integer, primary_key=True)
    keyword = Column(String(255))
    value = Column(String(255))
    user_id = Column(Integer, ForeignKey('user.id'), nullable=False)
    parent_id = Column(Integer, ForeignKey('address.id'), nullable=True)

    @classmethod
    def complex_cls_method(cls, user_id_, key_, value_):
        main = Address(keyword=key_, value="", user_id=user_id_, parent_id=None)
        session.add_all([main])
        session.flush()
        addrs = [Address(keyword=key_, value=item, user_id=user_id_, parent_id=main.id) for item in value_]
        session.add_all(addrs)
        return [main] + addrs


if __name__ == "__main__":
    # Model.metadata.create_all(engine)
    user = User([u"address1", u"address2"], "simple")
    session.add(user)
    session.flush()

    # as it can't be done in constructor, these additional statements needed 

    user.addresses = Address.complex_cls_method(
        user_id_=user.id,
        key_="address",
        value_=[u"address1", u"address2"]
    )
    session.commit()

问题是，是否有语法上优雅(并且技术上合理)的方法可以使用 User 的构造函数来执行此操作，或者在 session.flush 之后调用 User 类的单独方法来将所需的对象添加到关系中是否更安全(如示例代码)？

完全放弃构造函数仍然是可能的，但不太理想的选择，因为导致签名更改将需要大量重构。

Answer 1

您可以让 SQLAlchemy 处理持久化对象图，而不是手动刷新和设置 id 等。您只需要再一个adjacency list relationship在地址中，您就完成了:

class User(Model):
    __tablename__ = 'user'
    id = Column(Integer, primary_key=True)
    simple = Column(String(255))
    main_address = Column(String(255))
    addresses = relationship("Address",
                             cascade="all, delete-orphan")

    def __init__(self, addresses, simple):
        self.simple = simple
        self.main_address = addresses[0]
        self.addresses = Address.complex_cls_method(
            key="address",
            values=addresses
        )


class Address(Model):
    __tablename__ = 'address'
    id = Column(Integer, primary_key=True)
    keyword = Column(String(255))
    value = Column(String(255))
    user_id = Column(Integer, ForeignKey('user.id'), nullable=False)
    parent_id = Column(Integer, ForeignKey('address.id'), nullable=True)
    # For handling parent/child relationships in factory method
    parent = relationship("Address", remote_side=[id])

    @classmethod
    def complex_cls_method(cls, key, values):
        main = cls(keyword=key, value="")
        addrs = [cls(keyword=key, value=item, parent=main) for item in values]
        return [main] + addrs    

if __name__ == "__main__":
    user = User([u"address1", u"address2"], "simple")
    session.add(user)
    session.commit()
    print(user.addresses)

请注意，没有手动刷新等。SQLAlchemy 会根据对象关系自动计算出所需的插入顺序，以便可以遵守行之间的依赖关系。这是 Unit of Work 的一部分模式。