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,我有三张桌子:

阶级辩护:

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,以及指示一个特定用户(由user.sessionId与user.id一起识别)的订阅频道.

例如,假设我们有四个通道: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并且缺少订阅,把它留空.

我正在与SQLalchemy atm一起使用的数据库引擎.是sqlite3

我已经在这两天抓住了这个问题,我没有问题通过订阅表连接所有这三个,但是然后所有用户没有订阅的频道都被省略了.

我希望我能够充分描述我的问题,提前感谢.

编辑:管理以涉及子查询的略微笨重的方式解决这个问题:

# 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,所以答案仍然非常受欢迎.

选项1:

Subscription只是一个多对多的关系对象,我建议你把它建模为这样,而不是作为一个单独的类.请参阅配置多对多关系文档SQLAlchemy/declarative.

您使用测试代码建模成为:

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,User当channels要求时,每个只发出1个SELECT语句而不是1个.

选项2:

但是如果你想让你保持模型,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.