python - sqlalchemy 使用中间表加入自我（工作 sql，而不是 sqlalchemy）

Question

我正在尝试使用中间表执行多对多连接。（首先在 Items 中选择行，在 FK 上加入 Attribution，然后在 Attribution 中加入其他 FK 以从 Items 中获取更多信息）Schema 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'))

查询如下所示：

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()

这仅提供一排；输出：

upper_third 1

我通过 sqlite 使用的 sql 提供了预期的三行，如下所示：

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？

（生成sqlite数据库和测试查询的完整可运行代码如下）

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)

标签： pythonsqlitejoinsqlalchemy

Answer 1

简短的回答：SQLAlchemy 按预期工作。

关键的罪魁祸首是当您查询映射模型时，您将获得模型的实例作为结果。如果多次返回相同的模型，SA 将确保每个只返回一次，这就是为什么您只返回 1 行而不是预期的 3 行。

您可以通过您构造的 SQL ( select * ...) 与 SA 为您创建的 SQL ( select items.*; SA 不创建SQL 的区别来看到这*一点，但关键是它仅从items表中选择。

解决方案是将其他实体添加query(...)到要返回的对象中：

final_items = aliased(Items, name="FinalItems")
proj = (
    session
    .query(Items, Attribution, final_items)  # IMPORTANT !!!
    .join(Attribution, ( (Items.name=="upper_third") & (Attribution.link_id_m==Items.linkid) ))
    .join(final_items, final_items.linkid==Attribution.link_id_d)
)

运行如下：

for l in proj.all():
    print(l)
    # print(l.Items, l.Attribution, l.FinalItems)  # also can access models using names.

...将产生一个列表tuple(Items, Attribution, Items)：

(<Items(id=5, linkid=1, name='upper_third', project=1)>, <Attribution(id=1, link_id_d=2, link_id_m=1)>, <Items(id=2, linkid=2, name='lower_first', project=2)>)
(<Items(id=5, linkid=1, name='upper_third', project=1)>, <Attribution(id=1, link_id_d=2, link_id_m=1)>, <Items(id=4, linkid=2, name='lower_second', project=2)>)
(<Items(id=5, linkid=1, name='upper_third', project=1)>, <Attribution(id=1, link_id_d=2, link_id_m=1)>, <Items(id=6, linkid=2, name='lower_third', project=2)>)

如果您真的想像示例中那样返回列，则可以执行statement查询。下面的代码

for row in session.execute(proj.statement):
    print(row)

将返回：

(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)