Question

sqlalchemy中是否有办法在单个查询中关闭声明性的多态连接加载？大部分时间都很好，但我有：

class A(Base) : 
   discriminator = Column('type', mysql.INTEGER(1), index=True, nullable=False)
   __mapper_args__ = { 'polymorphic_on' : discriminator }
   id = Column(Integer, primary_key=True)
   p = Column(Integer)

class B(A) : 
   __mapper_args__ = { 'polymorphic_identity' : 0 }
   id = Column(Integer, primary_key=True)
   x = Column(Integer)

class C(A) : 
   __mapper_args__ = { 'polymorphic_identity' : 1 }
   id = Column(Integer, primary_key=True)
   y = Column(String)

我想进行查询，以便获得B.x＆gt;的所有A.ids。 10，如果A实际上是B，或者C.y =='blah'，如果A实际上是C，则全部按p排序。

为了迭代地进行，我只是从第一部分开始 - “如果A实际上是B，则获取所有A.id，其中B.x> 10。”所以我想我会从外部联接开始：

session.query(A.id).outerjoin((B, B.id == A.id)).filter(B.x > 10)

...除了似乎没有办法避免让外连接（（B，B.id == A.id））子句生成A中所有内容与子选择中B中所有内容的完全连接。如果B没有从A继承，那么这种情况就不会发生，所以我认为这是多态声明代码生成。有没有办法把它关掉？或者强迫外部连接做我想做的事情？

我想要的是这样的：

select a.id from A a left outer join B b on b.id == a.id where b.x > 10

但我会得到类似的东西：

select a.id from A a left outer join (select B.id, B.x, A.id from B inner join A on B.id == A.id)

......顺便说一句，如果不可能，那么后者效率会低于前者吗？ sql引擎实际上是执行内连接，还是会忽略它？

Answer 1

你应该使用with_polymorphic()而不是outerjoin（），这似乎会返回预期的结果：

session.query(A).with_polymorphic(B).filter(B.x > 10).all()
# BEGIN
# SELECT "A".type AS "A_type", "A".id AS "A_id", "A".p AS "A_p", "B".id AS "B_id", "B".x AS "B_x" 
# FROM "A" LEFT OUTER JOIN "B" ON "A".id = "B".id 
# WHERE "B".x > ?
# (10,)
# Col ('A_type', 'A_id', 'A_p', 'B_id', 'B_x')

与：相比：

session.query(A.id).outerjoin((B, B.id == A.id)).filter(B.x > 10)
# BEGIN
# SELECT "A".id AS "A_id" 
# FROM "A" LEFT OUTER JOIN (SELECT "A".type AS "A_type", "A".id AS "A_id", "A".p AS "A_p", "B".id AS "B_id", "B".x AS "B_x" 
# FROM "A" JOIN "B" ON "A".id = "B".id) AS anon_1 ON anon_1."A_id" = "A".id 
# WHERE anon_1."B_x" > ?
# (10,)
# Col ('A_id',)

我用来测试这个代码，以防任何人想要测试SQLAlchemy这个简洁的位：

#!/usr/bin/env python
import logging
from sqlalchemy import *
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker

Base = declarative_base()

class A(Base) :
   __mapper_args__ = { 'polymorphic_on' : discriminator }
   __tablename__ = 'A'

   id = Column(Integer, primary_key=True)
   discriminator = Column('type', Integer, index=True, nullable=False)
   p = Column(Integer)

class B(A) :
   __mapper_args__ = { 'polymorphic_identity' : 0 }
   __tablename__ = 'B'

   id = Column(Integer, ForeignKey('A.id'), primary_key=True)
   x = Column(Integer)

class C(A) :
   __mapper_args__ = { 'polymorphic_identity' : 1 }
   __tablename__ = 'C'

   id = Column(Integer, ForeignKey('A.id'), primary_key=True)
   y = Column(String)

meta = Base.metadata
meta.bind = create_engine('sqlite://')
meta.create_all()

Session = sessionmaker()
Session.configure(bind=meta.bind)
session = Session()

log = logging.getLogger('sqlalchemy')
log.addHandler(logging.StreamHandler())
log.setLevel(logging.DEBUG)

session.query(A.id).outerjoin((B, B.id == A.id)).filter(B.x > 10).all()
session.query(A).with_polymorphic(B).filter(B.x > 10).all()

我使用SQLAlchemy 0.6.4在Python 2.7上运行它。

Answer 2

您可以尝试单独为每个子类构建查询，然后将它们组合在一起。查询B.id时，SQLAlchemy隐式加入超类并返回A.id，因此对B.id和C.id的选择联合只返回一列。

>>> b_query = session.query(B.id).filter(B.x > 10)
>>> c_query = session.query(C.id).filter(C.y == 'foo')
>>> print b_query.union(c_query)
SELECT anon_1."A_id" AS "anon_1_A_id" 
FROM (SELECT "A".id AS "A_id" 
FROM "A" JOIN "B" ON "A".id = "B".id 
WHERE "B".x > ? UNION SELECT "A".id AS "A_id" 
FROM "A" JOIN "C" ON "A".id = "C".id 
WHERE "C".y = ?) AS anon_1

您仍然可以获得一个子选择，但只有一个“连接层” - 外部选择只是重命名该列。

sqlalchemy：关闭声明性多态连接？

2 个答案: