Python 编写引用同一表两次的SQL查询

考虑下表中的定义

class MCastSession(Base):
  __tablename__ = 'mcast_session'
  id = Column(Integer, primary_key=True)
  ip = Column(Integer)
  port = Column(Integer)
  __table_args__ = ( UniqueConstraint('ip', 'port'), )

class Topic(Base):
  __tablename__ = 'topic'
  id = Column(Integer, primary_key=True)
  name = Column(String, unique=True)
  mcast_session_id = Column(Integer, ForeignKey('mcast_session.id'))
  mcast_session = relationship('MCastSession')

class Host(Base):
  __tablename__ = 'host'
  id = Column(Integer, primary_key=True)
  name = Column(String, unique=True)

class Subscriber(Base):
  __tablename__ = 'subscriber'
  id = Column(Integer, primary_key=True)
  topic_id = Column(Integer, ForeignKey('topic.id'))
  topic = relationship('Topic')
  host_id = Column(Integer, ForeignKey('host.id'))
  host = relationship('Host')
  __table_args__ = ( UniqueConstraint('topic_id', 'host_id'), )

Example data:
Topic Session
T1    IP1:port1
T2    IP1:port2
T3    IP1:port2
T4    IP2:port1

Topic Host
T1    H1
T2    H1
T4    H2

---

我想编写一个查询，以获取订阅多播ip但不处理ip的所有主题的所有主机。在上面的例子中。H1有T1，因此订阅IP1，但没有T3，也有相同的IP1。因此查询应该返回H1。H2处理其订阅的IP（T4）的所有主题（T4），因此H2不应出现在结果中。如何为上述内容编写查询

下面的查询将生成目标：

select distinct host1.name host_name
  from Subscriber Subscriber1
 inner join host host1
    on host1.id = Subscriber1.Host_Id
 inner join topic topic1
    on topic1.id = Subscriber1.Topic_Id
 inner join mcast_session mcast_session1
    on mcast_session1.id = topic1.mcast_session_id
 where (select count(*)
          from mcast_session
         where mcast_session.ip = mcast_session1.ip) !=
       (select count(*)
          from topic
         inner join mcast_session
            on topic.mcast_session_id = mcast_session.id
         where mcast_session.ip = mcast_session1.ip)

对于逻辑的解释，查询可能有助于：

select  host1.name host_name,
       topic1.name topic_name,
       mcast_session1.ip,
       mcast_session1.port,
(select count(*)
          from mcast_session
         where mcast_session.ip = mcast_session1.ip) host_to_topic_registeration,
       (select count(*)
          from topic
         inner join mcast_session
            on topic.mcast_session_id = mcast_session.id
         where mcast_session.ip = mcast_session1.ip
           ) ip_topic_count
  from Subscriber Subscriber1
 inner join host host1
    on host1.id = Subscriber1.Host_Id
 inner join topic topic1
    on topic1.id = Subscriber1.Topic_Id
 inner join mcast_session mcast_session1
    on mcast_session1.id = topic1.mcast_session_id

---

：

查看所需结果的另一种方法是应用以下逻辑：

对于每个

IP

地址，计算

主题的总数

对于每个

主机

计算按

IP

地址分组的

主题

广播的总数

所需主机是指每个

IP

地址的

主题数不等于
IP
地址总数的主机

以下SA代码应为您提供所需的

主机

实例：

它生成下面的

SQL

代码（对于

SQLite

）：

---

现在，如果要在查询中多次使用同一个表，可以使用。下面的代码将返回元组列表

（mcasession，NNN）

，其中

NNN

是具有相同

IP

的MCastSession对象数：

aliased_MCastSession = aliased(MCastSession, name="MCastSession2")
qry = session.query(\
    MCastSession, \
    func.count(aliased_MCastSession.id).label("number_with_same_ip")).\
filter(MCastSession.ip == aliased_MCastSession.ip).\
group_by(MCastSession)

---

但是我不需要为提出的解决方案这样做，因为我使用了子查询。

如果有人可以帮助只使用原始SQL查询，这也将有助于两次引用同一个表：

从my_选项卡中选择一个*作为第一个_引用将my_选项卡作为第一个_引用的第二个_引用加入my_选项卡。ID=second_reference.ID+1

SELECT  host.id AS host_id, host.name AS host_name

FROM    host

JOIN   (SELECT  host.id AS host_id,
                mcast_session.ip AS mcast_session_ip,
                count(topic.id) AS num_topics
        FROM    host
        JOIN    subscriber
            ON  host.id = subscriber.host_id
        JOIN    topic
            ON  topic.id = subscriber.topic_id
        JOIN    mcast_session
            ON  mcast_session.id = topic.mcast_session_id
        GROUP BY host.id, mcast_session.ip
       ) AS host_ip_topics
    ON  host.id = host_ip_topics.host_id

JOIN   (SELECT  mcast_session.ip AS mcast_session_ip,
                count(topic.id) AS num_topics
        FROM    mcast_session
        JOIN    topic
            ON  mcast_session.id = topic.mcast_session_id
        GROUP BY mcast_session.ip
       ) AS ip_topics
    ON  host_ip_topics.mcast_session_ip = ip_topics.mcast_session_ip
    AND host_ip_topics.num_topics != ip_topics.num_topics

aliased_MCastSession = aliased(MCastSession, name="MCastSession2")
qry = session.query(\
    MCastSession, \
    func.count(aliased_MCastSession.id).label("number_with_same_ip")).\
filter(MCastSession.ip == aliased_MCastSession.ip).\
group_by(MCastSession)