如果我有模型:
class A(Base):
a_id = Column(Integer, primary_key=True)
other = Column(Text)
name = Column(Text)
class B(Base):
b_id = Column(Integer, primary_key=True)
other = Column(Text)
class C(Base):
c_id = Column(Integer, primary_key=True)
b_id = Column(ForeignKey(B.b_id))
b = relationship(B)
name = Column(Text)
a = relationship(
A,
primaryjoin=and_(
A.name == foreign(name),
A.other == foreign(B.other)))
如何获得关系 C.a
工作 session.query(C).options(joinedload(C.a))
不会失败。
在普通 SQL 中我会这样做:
select * from c
join b using(b_id)
join a on a.name = c.name and a.other = b.other
a.name, a.other
存在唯一约束,所以我知道我会得到 1 或 0 a
每c
.
我想我需要使用secondary=
在某种程度上,但我能找到的所有示例都是纯粹的多对多示例。
谢谢!
最佳答案
您必须使用 relationship to a non primary mapper 而不是辅助设备。 ,因为:
There is one complex join case where even this technique (composite "secondary" join) is not sufficient; when we seek to join from
A
toB
, making use of any number ofC
,D
, etc. in between, however there are also join conditions betweenA
andB
directly. In this case, the join fromA
toB
may be difficult to express with just a complexprimaryjoin
condition, as the intermediary tables may need special handling, and it is also not expressable with asecondary
object, since theA->secondary->B
pattern does not support any references betweenA
andB
directly.
因此,按照文档中的示例,我们可以使用 non primary mapper 建立您的关系。将类 A
映射到表 a
和 b
之间的联接:
from sqlalchemy import create_engine, Column, Integer, Text, ForeignKey, join, and_
from sqlalchemy.orm import relationship, mapper, sessionmaker, foreign
from sqlalchemy.ext.declarative import declarative_base
engine = create_engine('sqlite:///', echo=True)
Base = declarative_base()
Base.metadata.bind = engine
Session = sessionmaker()
class A(Base):
__tablename__ = 'a'
a_id = Column(Integer, primary_key=True)
other = Column(Text)
name = Column(Text)
class B(Base):
__tablename__ = 'b'
b_id = Column(Integer, primary_key=True)
other = Column(Text)
j = join(A, B, A.other == B.other)
A_viab = mapper(A, j, non_primary=True, properties={
"other": [j.c.a_other, j.c.b_other]
})
class C(Base):
__tablename__ = 'c'
c_id = Column(Integer, primary_key=True)
b_id = Column(ForeignKey(B.b_id))
b = relationship(B)
name = Column(Text)
a = relationship(
A_viab,
primaryjoin=and_(foreign(name) == A_viab.c.name,
b_id == A_viab.c.b_id))
实际操作:
In [4]: session.add(A(name='name', other='other'))
In [5]: session.add(C(name='name', b=B(other='other')))
In [6]: session.commit()
...
In [7]: c = session.query(C).options(joinedload(C.a)).first()
2018-01-04 15:10:21,338 INFO sqlalchemy.engine.base.Engine BEGIN (implicit)
2018-01-04 15:10:21,340 INFO sqlalchemy.engine.base.Engine SELECT c.c_id AS c_c_id, c.b_id AS c_b_id, c.name AS c_name, a_1.other AS a_1_other, b_1.other AS b_1_other, a_1.a_id AS a_1_a_id, a_1.name AS a_1_name, b_1.b_id AS b_1_b_id
FROM c LEFT OUTER JOIN (a AS a_1 JOIN b AS b_1 ON a_1.other = b_1.other) ON c.name = a_1.name AND c.b_id = b_1.b_id
LIMIT ? OFFSET ?
2018-01-04 15:10:21,340 INFO sqlalchemy.engine.base.Engine (1, 0)
In [8]: c.a
Out[8]: <cplxjoin.A at 0x7f0c81599630>
生成的查询与您的手动 SQL 并不完全相同,但 afaic 应该是等效的。
关于python - sqlalchemy 关系 PrimaryJoin 与自身列和父列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48080591/