使用ID而不是对象填充SQLAlchemy多对多关系 [英] Populating a SQLAlchemy many-to-many relationship using ID's instead of objects

问题描述

情况:

因此，我在SQLAlchemy中使用关联表.
例如，一个人可以参加许多聚会，而一个聚会可以有许多人作为客人:

So, I have a basic many-to-many relationship in SQLAlchemy using an association table.
For example, a person can attend many parties, and a party can have many persons as guests:

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，如果我想添加参加聚会的嘉宾列表，我会执行以下操作:

Normally if I wanted to add a list of guests to a party, I would do something like this:

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

......其中王子，奥利维亚和布列塔尼都是<Person>实例，而my_party是<SexyParty>实例.

...where prince, olivia and brittany are all <Person> instances, and my_party is a <SexyParty> instance.

我的问题:

我想使用个人ID而不是实例将客人添加到聚会中. 例如:

I'd like to add guests to a party using person ID's rather than instances. For example:

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，但这将导致不必要的DB查询，只是为了设置简单的多对多关系.

I could always load the instances from the databases, but that would entail an unnecessary DB query just to set a simple many-to-many relationships.

如何使用person_id的列表设置.guests属性? 必须使用一种简单的方法来完成此操作，因为关联表最终还是使用ID来表示多对多关系...

How would I go about setting the .guests attribute using a list of person_id's? There has to be a simple way to do this since the association table ultimately represents the many-to-many relationship using ID's anyway...

预先感谢，希望问题清楚.

thanks in advance, hope the question is clear.

推荐答案

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()

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