序列化嵌套查询结果

我在用烧瓶炼金术和棉花糖。我这里描述了一个简单的一对多关系(一 Broker 可以有很多 Account s) :

class Account(Base):
    __tablename__ = 'accounts'
    id = Column(Integer, primary_key=True)
    broker_id = Column(Integer, ForeignKey('brokers.id'))
    alias = Column(String)

    # relationships
    broker = relationship("Broker", back_populates="accounts", uselist=False)

class Broker(Model):
    __tablename__ = 'brokers'
    id = Column(Integer, primary_key=True)
    name = Column(String)

    # relationships
    accounts = relationship("Account", back_populates="broker")

我的棉花糖模式定义如下:

class AccountSchema(ma.SQLAlchemyAutoSchema):
    class Meta:
        table = Account.__table__
        sqla_session = db.session

account_schema = AccountSchema()
accounts_schema = AccountSchema(many=True)


class BrokerSchema(ma.SQLAlchemyAutoSchema):
    class Meta:
        table = Broker.__table__
        sqla_session = db.session

    accounts = ma.Nested(accounts_schema)

broker_schema = BrokerSchema()
brokers_schema = BrokerSchema(many=True)

我的 经纪人 路由可选择接受 broker_id 和 account_id . 看起来是这样的:

@api.route("")
@api.route("/<int:broker_id>")
@api.route("/<int:broker_id>/accounts")
@api.route("/<int:broker_id>/accounts/<int:account_id>")
class BrokerResource(Resource):
    def get(self, brokerage_id=None, account_id=None):
        # construct query
        query = ...

        return {
            "data": brokerages_schema.dump(query.all())
        }

我想创建一个查询,该查询将返回适合使用 BrokerSchema . 如果否 帐户\u id 已通过,则应返回所有子级 账户 物体。如果 帐户\u id 如果通过,则应返回单个记录。

如何构造这样的查询?