[英]Relationship spanning four tables in SQLAlchemy
我正在嘗試使關系跨越四個表。 我根據此問題中的代碼簡化了代碼,以匹配我的數據庫。
from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
class A(Base):
__tablename__ = 'a'
id = Column(Integer, primary_key=True)
b_id = Column(Integer, ForeignKey('b.id'))
# FIXME: This fails with:
# "Relationship A.ds could not determine any unambiguous local/remote column pairs based on
# join condition and remote_side arguments. Consider using the remote() annotation to
# accurately mark those elements of the join condition that are on the remote side of the relationship."
#
# ds = relationship("D", primaryjoin="and_(A.b_id == B.id, B.id == C.b_id, D.id == C.d_id)", viewonly=True)
def dq(self):
return sess.query(D).filter(and_(D.id == C.d_id,
C.b_id == B.id,
B.id == A.id,
A.id == self.id))
class B(Base):
__tablename__ = 'b'
id = Column(Integer, primary_key=True)
class C(Base):
__tablename__ = 'c'
b_id = Column(Integer, ForeignKey('b.id'), primary_key=True)
d_id = Column(Integer, ForeignKey('d.id'), primary_key=True)
class D(Base):
__tablename__ = 'd'
id = Column(Integer, primary_key=True)
e = create_engine("sqlite://", echo=True)
Base.metadata.create_all(e)
sess = Session(e)
sess.add(D(id=1))
sess.add(D(id=2))
sess.add(B(id=1))
sess.add(C(b_id=1, d_id=1))
sess.add(C(b_id=1, d_id=2))
sess.add(A(id=1, b_id=1))
sess.flush()
a1 = sess.query(A).first()
print a1.dq().all()
#print a1.ds
所以我的問題是'ds'關系的聯接語法。 當前的錯誤提到添加remote(),但是我還沒有使其工作。 我也嘗試使用secondaryjoin但沒有運氣。 “ dq”中的查詢有效,我最終可以通過在代碼中使用過濾器來解決該問題-我仍然很好奇如果可能的話,如何構建relathioship?
我不是sqlalchemy專家,這是我的理解。
我認為sqlalchemy的關系API中造成混亂的主要原因是,參數primaryjoin
, secondary
和secondaryjoin
真正含義是什么。 對我來說,它們是:
primaryjoin secondaryjoin(optional)
source -------------> secondary -------------------------> dest
(A) (D)
現在我們需要弄清楚中間部分應該是什么。 盡管自定義聯接在sqlalchemy中異常復雜,但您確實需要了解您要的內容,即原始SQL。 一種可能的解決方案是:
SELECT a.*, d.id
FROM a JOIN (b JOIN c ON c.b_id = b.id JOIN d ON d.id = c.d_id) /* secondary */
ON a.b_id = b.id /* primaryjoin */
WHERE a.id = 1;
在這種情況下,源a
與“輔助”聯接(b JOIN c .. JOIN d ..)
,並且由於D
已經存在於secondary
,因此沒有與D
的輔助(b JOIN c .. JOIN d ..)
。 我們有
ds1 = relationship(
'D',
primaryjoin='A.b_id == B.id',
secondary='join(B, C, B.id == C.b_id).join(D, C.d_id == D.id)',
viewonly=True, # almost always a better to add this
)
另一個解決方案可能是:
SELECT a.*, d.id
FROM a JOIN (b JOIN c ON c.b_id = b.id) /* secondary */
ON a.b_id = b.id /* primaryjoin */
JOIN d ON c.d_id = d.id /* secondaryjoin */
WHERE a.id = 1;
在這里, a
連接輔助c.d_id = d.id
(b JOIN c..)
,並且輔助c.d_id = d.id
d
的c.d_id = d.id
,因此:
ds2 = relationship(
'D',
primaryjoin='A.b_id == B.id',
secondary='join(B, C, B.id == C.b_id)',
secondaryjoin='C.d_id == D.id',
viewonly=True, # almost always a better to add this
)
經驗法則是將較長的聯接路徑放在輔助節點中,並將其鏈接到源和目標。
從性能角度ds1
, ds1
和ds2
導致查詢計划比dq
稍微簡單一些,但我認為它們之間沒有太大區別。 計划者總是知道得更多。
這是更新的代碼,供您參考。 請注意如何使用sess.query(A).options(joinedload('ds1'))
加載關系:
from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
class A(Base):
__tablename__ = 'a'
id = Column(Integer, primary_key=True)
b_id = Column(Integer, ForeignKey('b.id'))
ds1 = relationship(
'D',
primaryjoin='A.b_id == B.id',
secondary='join(B, C, B.id == C.b_id).join(D, C.d_id == D.id)',
viewonly=True, # almost always a better to add this
)
ds2 = relationship(
'D',
secondary='join(B, C, B.id == C.b_id)',
primaryjoin='A.b_id == B.id',
secondaryjoin='C.d_id == D.id',
viewonly=True, # almost always a better to add this
)
def dq(self):
return sess.query(D).filter(and_(D.id == C.d_id,
C.b_id == B.id,
B.id == A.id,
A.id == self.id))
class B(Base):
__tablename__ = 'b'
id = Column(Integer, primary_key=True)
class C(Base):
__tablename__ = 'c'
b_id = Column(Integer, ForeignKey('b.id'), primary_key=True)
d_id = Column(Integer, ForeignKey('d.id'), primary_key=True)
class D(Base):
__tablename__ = 'd'
id = Column(Integer, primary_key=True)
def __repr__(self):
return str(self.id)
e = create_engine("sqlite://", echo=True)
Base.metadata.drop_all(e)
Base.metadata.create_all(e)
sess = Session(e)
sess.add(D(id=1))
sess.add(D(id=2))
sess.add(B(id=1))
sess.add(B(id=2))
sess.flush()
sess.add(C(b_id=1, d_id=1))
sess.add(C(b_id=1, d_id=2))
sess.add(A(id=1, b_id=1))
sess.add(A(id=2, b_id=2))
sess.commit()
def get_ids(ds):
return {d.id for d in ds}
a1 = sess.query(A).options(joinedload('ds1')).filter_by(id=1).first()
print('{} a1.ds1: {}'.format('=' * 30, a1.ds1))
assert get_ids(a1.dq()) == get_ids(a1.ds1)
a1 = sess.query(A).options(joinedload('ds2')).filter_by(id=1).first()
print('{} a1.ds2: {}'.format('=' * 30, a1.ds2))
assert get_ids(a1.dq()) == get_ids(a1.ds2)
a2 = sess.query(A).options(joinedload('ds2')).filter_by(id=2).first()
print('{} a2.ds1: {}; a2.ds2 {};'.format('=' * 30, a2.ds1, a2.ds2))
assert a2.ds1 == a2.ds2 == []
對於查詢,您需要將表連接在一起。 即:
def db(self);
return sess.query(D).join(B).filter(...)
我不確定是否可以進行多個聯接,但是我不知道為什么不這樣做。 可以在這里找到對對象關系查詢的良好參考:
http://docs.sqlalchemy.org/en/rel_0_9/orm/tutorial.html#working-with-related-objects
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