Question

情况：

因此，我在SQLAlchemy中使用association table建立了基本的多对多关系例如，一个人可以参加许多聚会，一方可以有很多人作为嘉宾：

class Person(Base):
    __tablename__ = 'person'
    id      = Column(Integer, primary_key=True)
    name    = db.Column(db.String(50))


class SexyParty(Base):
    __tablename__ = 'sexy_party'
    id      = Column(Integer, primary_key=True)
    guests  = relationship('Person', secondary='guest_association',
                        lazy='dynamic', backref='parties')

guest_association = Table(
    'guest_association',
    Column('user_id',       Integer(), ForeignKey('person.id')),
    Column('sexyparty.id',  Integer(), ForeignKey('sexyparty.id'))
)

通常情况下，如果我想在聚会上添加一个客人列表，我会做这样的事情：

my_guests = [prince, olivia, brittany, me]
my_party.guests = guests
db.session.commit()

...其中prince，olivia和brittany都是<Person>个实例，而my_party是<SexyParty>个实例。

我的问题：

我想使用人ID而不是实例将来宾添加到派对中。例如：

guest_ids = [1, 2, 3, 5]
my_party.guests = guest_ids  # <-- This fails, because guest_ids
                             # are not <Person> instances

我总是可以从数据库中加载实例，但这需要一个不必要的数据库查询来设置一个简单的多对多关系。

如何使用person_id列表设置.guests属性？必须有一个简单的方法来做到这一点，因为关联表最终代表使用ID的多对多关系...

提前谢谢，希望问题很明确。

Answer 1

from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

class Person(Base):
    __tablename__ = 'person'
    id      = Column(Integer, primary_key=True)
    name    = Column(String(50))


class SexyParty(Base):
    __tablename__ = 'sexy_party'
    id      = Column(Integer, primary_key=True)
    guests  = relationship('Person', secondary='guest_association',
                        lazy='dynamic', backref='parties')

guest_association = Table(
    'guest_association', Base.metadata,
    Column('user_id',       Integer(), ForeignKey('person.id'), primary_key=True),
    Column('sexyparty_id',  Integer(), ForeignKey('sexy_party.id'), primary_key=True)
)


e = create_engine("sqlite://", echo=True)
Base.metadata.create_all(e)

sess = Session(e)

p1 = Person(id=1, name='p1')
p2 = Person(id=2, name='p2')
p3 = Person(id=3, name='p3')
p4 = Person(id=4, name='p4')

sp1 = SexyParty(id=1)

sess.add_all([sp1, p1, p2, p3, p4])
sess.commit()

# method one.  use insert()
sess.execute(guest_association.insert().values([(1, 1), (2, 1)]))

# method two.  map, optional association proxy
from sqlalchemy.ext.associationproxy import association_proxy

class GuestAssociation(Base):
    __table__ = guest_association
    party = relationship("SexyParty", backref="association_recs")

SexyParty.association_ids = association_proxy(
                    "association_recs", "user_id",
                    creator=lambda uid: GuestAssociation(user_id=uid))

sp1.association_ids.extend([3, 4])
sess.commit()

使用ID而不是对象填充SQLAlchemy多对多关系

1 个答案: