python - sqlalchemy:如何返回连接 3 个表的对象列表？

Question

@hybrid_method
# @paginate
def investors(self, **kwargs):
    """All investors for a given Custodian"""
    ind_inv_type_id = InvestorType.where(description="Individual").first().id
    inv_query = Investor.with_joined(InvestorAddress, InvestmentAddress, CustodianAddress) \
                        .filter_by(custodians_id=self.id) \
                        .with_joined(Investment) \
                        .filter_by(investor_types_id=ind_inv_type_id)
    investors = Investor.where(None, False, inv_query, **kwargs)

    temp_inv_query = Investor.with_joined(CustodianInvestor, Custodian)\
                             .filter_by(Custodian.id==self.id)
    temp_investors = Investor.where(None, False, temp_inv_query, **kwargs)
    return list(set(investors + temp_investors))
# end def investors



 # @auth.access_controlled
class InvestorAddress(db.Model, EntityAddressMixin):
    # Metadata
    __tablename__ = 'investor_addresses'

    # Database Columns
    investors_id = db.Column(db.ForeignKey("investors.investors_id"),
                             nullable=False)
    investor = db.relationship("Investor", foreign_keys=[investors_id],
                               backref=db.backref("InvestorAddress"))

# end class InvestorAddress

    class InvestmentAddress(db.Model):
    """This model differs from other EntityAddress Models because it links to either an investor_address or an custodian_address."""

    # Metadata
    __tablename__ = 'investment_addresses'

    # Database Columns
    address_types_id = db.Column(
        db.ForeignKey("address_types.address_types_id"),
        nullable=False)
    address_type = db.relationship("AddressType",
                                   foreign_keys=[address_types_id],
                                   backref=db.backref("InvestmentAddress"))

    investments_id = db.Column(db.ForeignKey("investments.investments_id"),
                               nullable=False)
    investment = db.relationship("Investment",
                                 foreign_keys=[investments_id],
                                 backref=db.backref("InvestmentAddress"))

    investor_addresses_id = db.Column(db.ForeignKey(
        "investor_addresses.investor_addresses_id"))
    investor_address = db.relationship("InvestorAddress",
                                       foreign_keys=[investor_addresses_id],
                                       backref=db.backref("InvestmentAddress"))

    custodian_addresses_id = db.Column(db.ForeignKey(
        "custodian_addresses.custodian_addresses_id"))
    custodian_address = db.relationship("CustodianAddress",
                                        foreign_keys=[custodian_addresses_id],
                                        backref=db.backref("InvestmentAddress")
                                        )
# end class InvestmentAddress



 class CustodianAddress(db.Model, EntityAddressMixin):
    """Defines the relationship between a Custodian and their addresses."""

    # Metadata
    __tablename__ = 'custodian_addresses'

    # Database Columns
    custodians_id = db.Column(db.ForeignKey(
        "custodians.custodians_id"), nullable=False)
    custodian = db.relationship("Custodian", foreign_keys=[custodians_id],
                                backref=db.backref("CustodianAddress"))

# end CustodianAddress

我有一个应用程序，此函数应该返回给定“托管人”的“投资者”列表。现在，当它执行时，我收到错误:“sqlalchemy.exc.ArgumentError:映射器选项需要字符串键或属性列表”。该错误来自“inv_query”中的“join”。 我已经包含了我用于连接的 3 个模型。

Answer 1

如您提供的文档中所述。 here

您应该在 with_joined 中提供字符串参数(表名称)。鉴于您已经定义了关系

Investor.with_joined('investorAddressTable', 'investmentAddressTable, 'custodianAddressTable')

如果你可以使用 session ，那么你可以直接查询 ORM 类

session.query(Investor).join(InvestorAddress).join(InvestmentAddress).join(CustodianAddress).all() # will assume you have set the foreign key properly