如何使用 SQLalchemy 连接三个表并将所有列保留在其中一个表中？

所以，我有三张表：

类定义：

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)

NOTE:我知道 user.username 应该是唯一的，需要解决这个问题，并且我不确定为什么 SQLalchemy 创建一些带有双引号的行名称。

我正在尝试想出一种方法来检索所有频道，以及指示某个特定用户（由 user.sessionId 和 user.id 一起标识）订阅了哪些频道。

例如，假设我们有四个通道：channel1、channel2、channel3、channel4；用户：user1；谁订阅了频道 1 和频道 4。对 user1 的查询将返回类似以下内容的内容：

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

这是最好的结果，但由于我完全不知道如何完成订阅列，所以我一直尝试在用户有订阅的行和缺少订阅的行中获取特定的用户 ID ，将其留空即可。

我与 SQLalchemy atm 一起使用的数据库引擎。是sqlite3

我已经为此绞尽脑汁两天了，我可以通过订阅表将这三个频道连接在一起，但随后用户没有订阅的所有频道都会被忽略。

我希望我能够充分描述我的问题，提前致谢。

EDIT：设法以一种稍微笨拙的方式解决这个问题，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，我将继续寻找更优雅的解决方案，因此仍然非常欢迎答案。

选项1：

Subscription只是一个多对多关系对象，我建议您将其建模为这样的对象，而不是作为一个单独的类。看配置多对多关系 http://www.sqlalchemy.org/docs/reference/ext/declarative.html#configuring-many-to-many-relationships的文档SQLAlchemy/declarative.

您使用测试代码进行建模变为：

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)

请注意使用eagerload，这将仅发出 1 个 SELECT 语句，而不是每个 1 个User when channels被要求。

选项2：

但是，如果您想保留模型并只是创建一个 SA 查询来根据您的要求提供列，那么以下查询应该可以完成这项工作：

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。