python - sqlalchemy多对多，但是相反吗？

抱歉，如果不是“inverse”不是首选术语，那可能会妨碍我的搜索。无论如何，我要处理两个sqlalchemy声明性类，这是一个多对多关系。第一个是Account，第二个是Collection。用户“购买”了集合，但是我想显示用户尚未购买的前10个集合。

from sqlalchemy import *
from sqlalchemy.orm import scoped_session, sessionmaker, relation
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

engine = create_engine('sqlite:///:memory:', echo=True)
Session = sessionmaker(bind=engine)

account_to_collection_map = Table('account_to_collection_map', Base.metadata,
                                Column('account_id', Integer, ForeignKey('account.id')),
                                Column('collection_id', Integer, ForeignKey('collection.id')))

class Account(Base):
    __tablename__ = 'account'

    id = Column(Integer, primary_key=True)
    email = Column(String)

    collections = relation("Collection", secondary=account_to_collection_map)

    # use only for querying?
    dyn_coll = relation("Collection", secondary=account_to_collection_map, lazy='dynamic')

    def __init__(self, email):
        self.email = email

    def __repr__(self):
        return "<Acc(id=%s email=%s)>" % (self.id, self.email)

class Collection(Base):
    __tablename__ = 'collection'

    id = Column(Integer, primary_key=True)
    slug = Column(String)

    def __init__(self, slug):
        self.slug = slug

    def __repr__(self):
        return "<Coll(id=%s slug=%s)>" % (self.id, self.slug)

因此，使用account.collections可以获得所有集合，并且使用dyn_coll.limit(1).all()可以将查询应用于集合列表...但是我该如何做逆运算呢？我想获取该帐户而不是映射的前10个收藏集。

任何帮助都令人感激。谢谢!

最佳答案

我不会为此目的使用该关系，因为从技术上讲，它不是您正在建立的关系(因此，在两侧保持同步等所有技巧都将失效)。
IMO，最干净的方法是定义一个简单的查询，该查询将返回您要查找的对象:

class Account(Base):
    ...
    # please note added *backref*, which is needed to build the
    #query in Account.get_other_collections(...)
    collections = relation("Collection", secondary=account_to_collection_map, backref="accounts")

    def get_other_collections(self, maxrows=None):
        """ Returns the collections this Account does not have yet.  """
        q = Session.object_session(self).query(Collection)
        q = q.filter(~Collection.accounts.any(id=self.id))
        # note: you might also want to order the results
        return q[:maxrows] if maxrows else q.all()
...

关于python - sqlalchemy多对多，但是相反吗？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/3983593/