Question

我有以下型号：

class business_member(base_model):
    __tablename__ = "business_member"

    id = Column(Integer, primary_key=True)
    business_id = Column(Integer, ForeignKey("business.id"), nullable=False)
    member_id = Column(Integer, ForeignKey("user.id"), nullable=False)
    business_parts_id = Column(Integer, ForeignKey("business_parts.id"), nullable=False)
    is_admin = Column(Boolean, nullable=True, default=False)
    created_date = Column(DateTime, nullable=False)
    modified_date = Column(DateTime, nullable=True)

    __table_args__ = (UniqueConstraint("business_id", "member_id", name="business_member_uc"), )
    business = relationship("business", back_populates="members")
    member = relationship("user", back_populates="as_employee")
    part = relationship("business_parts", back_populates="employees")

我转换了以下原始SQL：

SELECT id, member_id, is_admin FROM business_member WHERE id IN (
SELECT id FROM business WHERE business_id = (
SELECT business_id FROM business_member WHERE member_id = 5 and business_id = 1))

到SQLAlchemy查询：

session = database.get_session()

user_validation_query = session.query(business_member.business_id).filter_by(
    member_id=kwargs["user_id"], 
    business_id=request.raw_args["business_id"]
).subquery()

get_business_id_query = session.query(business_member).options(
    load_only(business_member.id)
).filter(business_member.business_id == user_validation_query).subquery()

member_list = session.query(business_member).filter(
    business_member.id.in_(get_business_id_query)
).all()

一切正常，但我想在查询结果中返回id，is_admin和member（relationship）列，我尝试了不同的方法，如load_only()，但没有一种方法正常工作。我怎么能这样做？？

更新

例如，我将查询更改为：

member_list = session.query(business_member).options(
    load_only("id", "is_admin", "member")
).filter(
    business_member.id.in_(get_business_id_query)
).all()

我得到以下异常： can't locate strategy for (('deferred', False), ('instrument', True))

Answer 1

使用此

member_list = session.query(business_member.id, business_member.is_admin, business_member.member).filter( business_member.id.in_(get_business_id_query)).all()

在SQLAlchemy查询中选择关系

1 个答案: