Question

所以，我有三张桌子：

班级辩护：

engine = create_engine('sqlite://test.db', echo=False)
SQLSession = sessionmaker(bind=engine)
Base = declarative_base()

class Channel(Base):
    __tablename__ = 'channel'

    id = Column(Integer, primary_key = True)
    title = Column(String)
    description = Column(String)
    link = Column(String)
    pubDate = Column(DateTime)

class User(Base):
    __tablename__ = 'user'

    id = Column(Integer, primary_key = True)
    username = Column(String)
    password = Column(String)
    sessionId = Column(String)

class Subscription(Base):
    __tablename__ = 'subscription'

    userId = Column(Integer, ForeignKey('user.id'), primary_key=True)
    channelId = Column(Integer, ForeignKey('channel.id'), primary_key=True)

注意：我知道user.username应该是唯一的，需要修复它，我不确定为什么SQLalchemy会使用双引号创建一些行名称。

我正在尝试找到一种方法来检索所有频道，以及指示某个特定用户（由user.sessionId与user.id一起识别）在哪些频道上订阅。< / p>

例如，假设我们有四个频道：channel1，channel2，channel3，channel4;用户：user1;谁在channel1和channel4上订阅。 user1的查询将返回如下内容：

channel.id | channel.title | subscribed
---------------------------------------
1            channel1        True
2            channel2        False
3            channel3        False
4            channel4        True

这是一个最好的结果，但由于我完全不知道如何完成订阅列，我一直试图在用户订阅的行中获取特定用户ID，并且订阅缺失，请留空。

我正在与SQLalchemy atm一起使用的数据库引擎。是sqlite3

我现在已经对此感到头疼了两天，我通过订阅表将所有这三个连接在一起没有问题，但是用户没有订阅的所有频道都被省略了。< / p>

我希望我能够充分描述我的问题，提前谢谢。

编辑：管理以稍微笨重的方式解决此问题涉及子查询：

# What a messy SQL query!
stmt = query(Subscription).filter_by(userId = uid()).join((User, Subscription.userId == User.id)).filter_by(sessionId = id()).subquery()
subs = aliased(Subscription, stmt)
results = query(Channel.id, Channel.title, subs.userId).outerjoin((subs, subs.channelId == Channel.id))

但是，我会继续寻找更优雅的解决方案，所以答案仍然非常受欢迎。

Answer 1

方法-1：

Subscription只是一个多对多关系对象，我建议你将其建模为一个单独的类而不是一个单独的类。请参阅SQLAlchemy/declarative的{{3}}文档。

您使用测试代码建模：

from sqlalchemy import create_engine, Column, Integer, DateTime, String, ForeignKey, Table from sqlalchemy.orm import relation, scoped_session, sessionmaker, eagerload from sqlalchemy.ext.declarative import declarative_base engine = create_engine('sqlite:///:memory:', echo=True) session = scoped_session(sessionmaker(bind=engine, autoflush=True)) Base = declarative_base() t_subscription = Table('subscription', Base.metadata, Column('userId', Integer, ForeignKey('user.id')), Column('channelId', Integer, ForeignKey('channel.id')), ) class Channel(Base): __tablename__ = 'channel' id = Column(Integer, primary_key = True) title = Column(String) description = Column(String) link = Column(String) pubDate = Column(DateTime) class User(Base): __tablename__ = 'user' id = Column(Integer, primary_key = True) username = Column(String) password = Column(String) sessionId = Column(String) channels = relation("Channel", secondary=t_subscription) # NOTE: no need for this class # class Subscription(Base): # ... Base.metadata.create_all(engine) # ###################### # Add test data c1 = Channel() c1.title = 'channel-1' c2 = Channel() c2.title = 'channel-2' c3 = Channel() c3.title = 'channel-3' c4 = Channel() c4.title = 'channel-4' session.add(c1) session.add(c2) session.add(c3) session.add(c4) u1 = User() u1.username ='user1' session.add(u1) u1.channels.append(c1) u1.channels.append(c3) u2 = User() u2.username ='user2' session.add(u2) u2.channels.append(c2) session.commit() # ###################### # clean the session and test the code session.expunge_all() # retrieve all (I assume those are not that many) channels = session.query(Channel).all() # get subscription info for the user #q = session.query(User) # use eagerload(...) so that all 'subscription' table data is loaded with the user itself, and not as a separate query q = session.query(User).options(eagerload(User.channels)) for u in q.all(): for c in channels: print (c.id, c.title, (c in u.channels))

产生以下输出：

(1, u'channel-1', True) (2, u'channel-2', False) (3, u'channel-3', True) (4, u'channel-4', False) (1, u'channel-1', False) (2, u'channel-2', True) (3, u'channel-3', False) (4, u'channel-4', False)

请注意eagerload的使用，当{1}}被要求时，User只会为每个channels发出1个SELECT语句而不是1个。

方法-2：

但是如果你想让你保持模型并且只是创建一个SA查询，可以根据你的要求提供列，那么下面的查询就可以完成这项工作：

from sqlalchemy import and_ from sqlalchemy.sql.expression import case #... q = (session.query(#User.username, Channel.id, Channel.title, case([(Subscription.channelId == None, False)], else_=True) ).outerjoin((Subscription, and_(Subscription.userId==User.id, Subscription.channelId==Channel.id)) ) ) # optionally filter by user q = q.filter(User.id == uid()) # assuming uid() is the function that provides user.id q = q.filter(User.sessionId == id()) # assuming uid() is the function that provides user.sessionId res = q.all() for r in res: print r

输出与上面的选项-1完全相同。

Answer 2

为了使这个更容易，我已经为你的模型添加了关系，这样你就可以用user.subscriptions来获得所有的订阅。

engine = create_engine('sqlite://test.db', echo=False)
SQLSession = sessionmaker(bind=engine)
Base = declarative_base()

class Channel(Base):
    __tablename__ = 'channel'

    id = Column(Integer, primary_key = True)
    title = Column(String)
    description = Column(String)
    link = Column(String)
    pubDate = Column(DateTime)

class User(Base):
    __tablename__ = 'user'

    id = Column(Integer, primary_key = True)
    username = Column(String)
    password = Column(String)
    sessionId = Column(String)

class Subscription(Base):
    __tablename__ = 'subscription'

    userId = Column(Integer, ForeignKey('user.id'), primary_key=True)
    user = relationship(User, primaryjoin=userId == User.id, backref='subscriptions')
    channelId = Column(Integer, ForeignKey('channel.id'), primary_key=True)
    channel = relationship(channel, primaryjoin=channelId == channel.id, backref='subscriptions')

results = session.query(
    Channel.id,
    Channel.title,
    Channel.subscriptions.any().label('subscribed'),
)

for channel in results:
    print channel.id, channel.title, channel.subscribed

Answer 3

不要向用户查询。来自频道的查询。

user = query(User).filter_by(id=1).one()
for channel in query(Channel).all():
    print channel.id, channel.title, user in channel.subscriptions.user

通过这种方式，您可以获得所有渠道，而不仅仅是与相关用户相关的渠道。

如何使用SQLalchemy连接三个表并将所有列保留在其中一个表中？

3 个答案: