[sqlalchemy通过中间表联接到自身（正在运行的sql，不是sqlalchemy）]]

Question

我正在尝试与一个中间表进行多对多联接。 （首先在Items中选择行，在FK上加入归因，然后在归因中与另一个FK结合以从Items中获得更多信息）模式如下所示：

class Items(Base):
    __tablename__ = "items"

    id = Column(Integer, primary_key=True)
    name = Column(Text)
    project = Column(Integer)
    linkid = Column(Integer, ForeignKey("items.linkid"))

    linked = relationship("Items", foreign_keys="[Items.linkid]")

class Attribution(Base):
    __tablename__ = "attribution"

    id = Column(Integer, primary_key=True)
    link_id_d = Column(Integer, ForeignKey('items.linkid'))
    link_id_m = Column(Integer, ForeignKey('items.linkid'))
查询看起来像这样：

final_items = aliased(Items)
proj_1 = session.query(Items)\
    .join(Attribution, ( (Items.name=="upper_third") & (Attribution.link_id_m==Items.linkid) ))\
    .join(final_items, final_items.linkid==Attribution.link_id_d)\
    .all()
这仅提供一行；输出：

upper_third 1
我正在通过sqlite使用的sql，它提供了预期的三行，看起来像这样：

SELECT * FROM items
join attribution on items.name = 'upper_third' and items.linkid == attribution.link_id_m
join items as tbl1 on tbl1.linkid == attribution.link_id_d
输出：

5|upper_third|1|1|1|2|1|2|lower_first|2|2
5|upper_third|1|1|1|2|1|4|lower_second|2|2
5|upper_third|1|1|1|2|1|6|lower_third|2|2
这两个查询之间的逻辑区别是什么，如何将sql解决方案迁移到sqalchemy？

（用于生成sqlite数据库和测试查询的完整可运行代码在下面]

from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import create_engine, UniqueConstraint, ForeignKey, func, and_
from sqlalchemy.orm import sessionmaker, relationship, aliased, backref

Base = declarative_base()

from sqlalchemy import Column, Integer, String, Float, Boolean, Text, Table

class Items(Base):
    __tablename__ = "items"

    id = Column(Integer, primary_key=True)
    name = Column(Text)
    project = Column(Integer)
    linkid = Column(Integer, ForeignKey("items.linkid"))

    linked = relationship("Items", foreign_keys="[Items.linkid]")
    attributions_m = relationship("Attribution", foreign_keys="[Attribution.link_id_m]")
    attributions_d = relationship("Attribution", foreign_keys="[Attribution.link_id_d]")

class Attribution(Base):
    __tablename__ = "attribution"

    id = Column(Integer, primary_key=True)
    link_id_d = Column(Integer, ForeignKey('items.linkid'))  # one of the tests to link to
    link_id_m = Column(Integer, ForeignKey('items.linkid'))  # one of the tests to link to


import os


if os.path.exists('app.db'):
    os.remove('app.db')

engine = create_engine('sqlite+pysqlite:///app.db')

Base.metadata.create_all(engine)

Session = sessionmaker(bind=engine)
session = Session()


u1 = Items(name="upper_first", linkid=1, project=1)
l1 = Items(name="lower_first", linkid=2, project=2)

u2 = Items(name="upper_second", linkid=1, project=1)
l2 = Items(name="lower_second", linkid=2, project=2)

u3 = Items(name="upper_third", linkid=1, project=1)
l3 = Items(name="lower_third", linkid=2, project=2)

session.add(u1)
session.add(l1)
session.add(u2)
session.add(l2)
session.add(u3)
session.add(l3)

session.commit()

a1 = Attribution(link_id_m=u3.linkid, link_id_d=l3.linkid)
session.add(a1)
session.commit()

final_items = aliased(Items)
proj = session.query(Items)\
    .join(Attribution, ( (Items.name=="upper_third") & (Attribution.link_id_m==Items.linkid) ))\
    .join(final_items, final_items.linkid==Attribution.link_id_d)\
    .all()

for l in proj:
    print(l.name, l.linkid)

我正在尝试与一个中间表进行多对多联接。 （首先选择“项目”中的行，然后加入FK上的归因，然后与其他FK一起归因以从“项目”中获取更多信息）...

Answer 1

简短的答案：SQLAlchemy正常工作。