具有关联对象的多对多和定义的所有关系在删除时崩溃
Many-to-Many with association object and all relationships defined crashes on delete
当具有描述了所有关系的完全成熟的多对多时,删除两个主要对象之一会崩溃。
描述
汽车 (.car_ownerships
) <-> (.car
) 汽车所有权 (.person
) <-> (.car_ownerships
) 人
汽车 (.people
) <----------------> (.cars
) 人
问题
删除汽车或人时
SA首先删除关联对象CarOwnership(因为'through'与secondary
参数的关系),然后尝试将同一个关联中的外键更新为NULL对象,因此崩溃。
我该如何解决?我有点困惑地看到文档中没有解决这个问题,也没有在我可以在网上找到的任何地方解决,因为我认为这种模式很常见:-/。我错过了什么?
我知道我可以为直通关系打开 passive_deletes
开关,但我想保留删除语句,只是为了防止更新发生或(让它发生之前)。
编辑:实际上,如果session中加载了依赖对象,passive_deletes
并不能解决问题,因为DELETE
语句仍然会被发出.一个解决方案是使用 viewonly=True
,但我不仅失去了删除,而且失去了关联对象的自动创建。另外我觉得 viewonly=True
很危险,因为它让你 append()
不用坚持!
REPEX
设置
from sqlalchemy import create_engine, Table, Column, Integer, String, ForeignKey
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship, backref, sessionmaker
engine = create_engine('sqlite:///:memory:', echo = False)
Base = declarative_base()
Session = sessionmaker(bind=engine)
session = Session()
class Person(Base):
__tablename__ = 'persons'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
cars = relationship('Car', secondary='car_ownerships', backref='people')
def __repr__(self):
return '<Person {} [{}]>'.format(self.name, self.id)
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
def __repr__(self):
return '<Car {} [{}]>'.format(self.name, self.id)
class CarOwnership(Base):
__tablename__ = 'car_ownerships'
id = Column(Integer(), primary_key=True)
type = Column(String(255))
car_id = Column(Integer(), ForeignKey(Car.id))
car = relationship('Car', backref='car_ownerships')
person_id = Column(Integer(), ForeignKey(Person.id))
person = relationship('Person', backref='car_ownerships')
def __repr__(self):
return 'Ownership [{}]: {} <<-{}->> {}'.format(self.id, self.car, self.type, self.person)
Base.metadata.create_all(engine)
归档对象
antoine = Person(name='Antoine')
rob = Person(name='Rob')
car1 = Car(name="Honda Civic")
car2 = Car(name='Renault Espace')
CarOwnership(person=antoine, car=car1, type = "secondary")
CarOwnership(person=antoine, car=car2, type = "primary")
CarOwnership(person=rob, car=car1, type = "primary")
session.add(antoine)
session.commit()
session.query(CarOwnership).all()
正在删除 -> 崩溃
print('#### DELETING')
session.delete(car1)
print('#### COMMITING')
session.commit()
# StaleDataError Traceback (most recent call last)
# <ipython-input-6-80498b2f20a3> in <module>()
# 1 session.delete(car1)
# ----> 2 session.commit()
# ...
诊断
我在上面提出的解释得到了引擎给出的 SQL 语句的支持 echo=True
:
#### DELETING
#### COMMITING
2016-07-07 16:55:28,893 INFO sqlalchemy.engine.base.Engine SELECT persons.id AS persons_id, persons.name AS persons_name
FROM persons, car_ownerships
WHERE ? = car_ownerships.car_id AND persons.id = car_ownerships.person_id
2016-07-07 16:55:28,894 INFO sqlalchemy.engine.base.Engine (1,)
2016-07-07 16:55:28,895 INFO sqlalchemy.engine.base.Engine SELECT car_ownerships.id AS car_ownerships_id, car_ownerships.type AS car_ownerships_type, car_ownerships.car_id AS car_ownerships_car_id, car_ownerships.person_id AS car_ownerships_person_id
FROM car_ownerships
WHERE ? = car_ownerships.car_id
2016-07-07 16:55:28,896 INFO sqlalchemy.engine.base.Engine (1,)
2016-07-07 16:55:28,898 INFO sqlalchemy.engine.base.Engine DELETE FROM car_ownerships WHERE car_ownerships.car_id = ? AND car_ownerships.person_id = ?
2016-07-07 16:55:28,898 INFO sqlalchemy.engine.base.Engine ((1, 1), (1, 2))
2016-07-07 16:55:28,900 INFO sqlalchemy.engine.base.Engine UPDATE car_ownerships SET car_id=? WHERE car_ownerships.id = ?
2016-07-07 16:55:28,900 INFO sqlalchemy.engine.base.Engine ((None, 1), (None, 2))
2016-07-07 16:55:28,901 INFO sqlalchemy.engine.base.Engine ROLLBACK
编辑
使用association_proxy
我们可以使用关联代理来尝试实现 'through' 关系。
然而,为了直接.append()
一个依赖对象,我们需要为关联对象创建一个构造函数。此构造函数必须 'hacked' 才能成为双向的,因此我们可以同时使用两个赋值:
my_car.people.append(Person(name='my_son'))
my_husband.cars.append(Car(name='new_shiny_car'))
生成的(经过中间测试的)代码如下,但我对它不太满意(因为这个 hacky 构造函数还有什么会破坏?)。
编辑: 下面的 RazerM 的回答中介绍了使用关联代理的方法。 association_proxy()
有一个 creator 参数,它减轻了我最终在下面使用的巨大构造函数的需要。
class Person(Base):
__tablename__ = 'persons'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
cars = association_proxy('car_ownerships', 'car')
def __repr__(self):
return '<Person {} [{}]>'.format(self.name, self.id)
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
people = association_proxy('car_ownerships', 'person')
def __repr__(self):
return '<Car {} [{}]>'.format(self.name, self.id)
class CarOwnership(Base):
__tablename__ = 'car_ownerships'
id = Column(Integer(), primary_key=True)
type = Column(String(255))
car_id = Column(Integer(), ForeignKey(Car.id))
car = relationship('Car', backref='car_ownerships')
person_id = Column(Integer(), ForeignKey(Person.id))
person = relationship('Person', backref='car_ownerships')
def __init__(self, car=None, person=None, type='secondary'):
if isinstance(car, Person):
car, person = person, car
self.car = car
self.person = person
self.type = type
def __repr__(self):
return 'Ownership [{}]: {} <<-{}->> {}'.format(self.id, self.car, self.type, self.person)
您使用的是 Association Object,因此您需要采取不同的方式。
我改变了这里的关系,仔细看它们,因为一开始你有点难以理解(至少对我来说是这样!)。
我使用了 back_populates
,因为在这种情况下它比 backref
更清晰。多对多关系的双方都必须直接引用 CarOwnership
,因为这是您要使用的对象。这也是您的示例显示的内容;您需要使用它,以便您可以设置 type
.
class Person(Base):
__tablename__ = 'persons'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
cars = relationship('CarOwnership', back_populates='person')
def __repr__(self):
return '<Person {} [{}]>'.format(self.name, self.id)
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
people = relationship('CarOwnership', back_populates='car')
def __repr__(self):
return '<Car {} [{}]>'.format(self.name, self.id)
class CarOwnership(Base):
__tablename__ = 'car_ownerships'
id = Column(Integer(), primary_key=True)
type = Column(String(255))
car_id = Column(Integer(), ForeignKey(Car.id))
person_id = Column(Integer(), ForeignKey(Person.id))
car = relationship('Car', back_populates='people')
person = relationship('Person', back_populates='cars')
def __repr__(self):
return 'Ownership [{}]: {} <<-{}->> {}'.format(self.id, self.car, self.type, self.person)
请注意,删除任一侧后,car_ownerships
行不会被删除,它只会将外键设置为 NULL。如果您想设置自动删除,我可以在我的答案中添加更多内容。
编辑:直接访问Car
和Person
对象的集合,需要使用association_proxy
,类 然后改成这样:
from sqlalchemy.ext.associationproxy import association_proxy
class Person(Base):
__tablename__ = 'persons'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
cars = association_proxy(
'cars_association', 'car', creator=lambda c: CarOwnership(car=c))
def __repr__(self):
return '<Person {} [{}]>'.format(self.name, self.id)
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
people = association_proxy(
'people_association', 'person', creator=lambda p: CarOwnership(person=p))
def __repr__(self):
return '<Car {} [{}]>'.format(self.name, self.id)
class CarOwnership(Base):
__tablename__ = 'car_ownerships'
id = Column(Integer(), primary_key=True)
type = Column(String(255), default='secondary')
car_id = Column(Integer(), ForeignKey(Car.id))
person_id = Column(Integer(), ForeignKey(Person.id))
car = relationship('Car', backref='people_association')
person = relationship('Person', backref='cars_association')
def __repr__(self):
return 'Ownership [{}]: {} <<-{}->> {}'.format(self.id, self.car, self.type, self.person)
编辑: 在您的编辑中,您在将其转换为使用 backref
时犯了一个错误。您的汽车和人的关联代理不能同时使用 'car_ownerships' 关系,这就是为什么我有一个名为 'people_association' 和一个名为 'cars_association'.
的原因
您拥有的 'car_ownerships' 关系与关联 table 被称为 'car_ownerships' 这一事实无关,因此我以不同的方式命名它们。
我修改了上面的代码块。要允许附加工作,您需要将创建者添加到关联代理。我已将 back_populates
更改为 backref
,并将默认 type
添加到 Column
对象而不是构造函数。
最干净的解决方案如下,不涉及关联代理。它是完全成熟的多对多关系所缺少的秘诀。
在这里,我们编辑从依赖对象 Car 和 Person 到关联对象 的直接关系CarOwnership,为了防止这些关系在关联对象被删除后发出UPDATE
。为此,我们使用 passive_deletes='all'
标志。
产生的互动是:
- 能够从依赖对象查询和设置关联对象
# Changing Ownership type:
my_car.car_ownerships[0].type = 'primary'
# Creating an ownership between a car and a person directly:
CarOwnership(car=my_car, person=my_husband, type='primary')
直接访问和编辑依赖对象的能力:
# Get all cars from a person:
[print(c) for c in my_husband.cars]
# Update the name of one of my cars:
me.cars[0].name = me.cars[0].name + ' Cabriolet'
创建或删除依赖对象时自动创建和删除关联对象
# Create a new owner and assign it to a car:
my_car.people.append(Person('my_husband'))
session.add(my_car)
session.commit() # Creates the necessary CarOwnership
# Delete a car:
session.delete(my_car)
session.commit() # Deletes all the related CarOwnership objects
代码
class Person(Base):
__tablename__ = 'persons'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
cars = relationship('Car', secondary='car_ownerships', backref='people')
def __repr__(self):
return '<Person {} [{}]>'.format(self.name, self.id)
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
def __repr__(self):
return '<Car {} [{}]>'.format(self.name, self.id)
class CarOwnership(Base):
__tablename__ = 'car_ownerships'
id = Column(Integer(), primary_key=True)
type = Column(String(255))
car_id = Column(Integer(), ForeignKey(Car.id))
car = relationship('Car', backref=backref('car_ownerships', passive_deletes='all'))
person_id = Column(Integer(), ForeignKey(Person.id))
person = relationship('Person', backref=backref('car_ownerships', passive_deletes='all'))
def __repr__(self):
return 'Ownership [{}]: {} <<-{}->> {}'.format(self.id, self.car, self.type, self.person)
当具有描述了所有关系的完全成熟的多对多时,删除两个主要对象之一会崩溃。
描述
汽车 (.car_ownerships
) <-> (.car
) 汽车所有权 (.person
) <-> (.car_ownerships
) 人
汽车 (.people
) <----------------> (.cars
) 人
问题
删除汽车或人时
SA首先删除关联对象CarOwnership(因为'through'与secondary
参数的关系),然后尝试将同一个关联中的外键更新为NULL对象,因此崩溃。
我该如何解决?我有点困惑地看到文档中没有解决这个问题,也没有在我可以在网上找到的任何地方解决,因为我认为这种模式很常见:-/。我错过了什么?
我知道我可以为直通关系打开 passive_deletes
开关,但我想保留删除语句,只是为了防止更新发生或(让它发生之前)。
编辑:实际上,如果session中加载了依赖对象,passive_deletes
并不能解决问题,因为DELETE
语句仍然会被发出.一个解决方案是使用 viewonly=True
,但我不仅失去了删除,而且失去了关联对象的自动创建。另外我觉得 viewonly=True
很危险,因为它让你 append()
不用坚持!
REPEX
设置
from sqlalchemy import create_engine, Table, Column, Integer, String, ForeignKey
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship, backref, sessionmaker
engine = create_engine('sqlite:///:memory:', echo = False)
Base = declarative_base()
Session = sessionmaker(bind=engine)
session = Session()
class Person(Base):
__tablename__ = 'persons'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
cars = relationship('Car', secondary='car_ownerships', backref='people')
def __repr__(self):
return '<Person {} [{}]>'.format(self.name, self.id)
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
def __repr__(self):
return '<Car {} [{}]>'.format(self.name, self.id)
class CarOwnership(Base):
__tablename__ = 'car_ownerships'
id = Column(Integer(), primary_key=True)
type = Column(String(255))
car_id = Column(Integer(), ForeignKey(Car.id))
car = relationship('Car', backref='car_ownerships')
person_id = Column(Integer(), ForeignKey(Person.id))
person = relationship('Person', backref='car_ownerships')
def __repr__(self):
return 'Ownership [{}]: {} <<-{}->> {}'.format(self.id, self.car, self.type, self.person)
Base.metadata.create_all(engine)
归档对象
antoine = Person(name='Antoine')
rob = Person(name='Rob')
car1 = Car(name="Honda Civic")
car2 = Car(name='Renault Espace')
CarOwnership(person=antoine, car=car1, type = "secondary")
CarOwnership(person=antoine, car=car2, type = "primary")
CarOwnership(person=rob, car=car1, type = "primary")
session.add(antoine)
session.commit()
session.query(CarOwnership).all()
正在删除 -> 崩溃
print('#### DELETING')
session.delete(car1)
print('#### COMMITING')
session.commit()
# StaleDataError Traceback (most recent call last)
# <ipython-input-6-80498b2f20a3> in <module>()
# 1 session.delete(car1)
# ----> 2 session.commit()
# ...
诊断
我在上面提出的解释得到了引擎给出的 SQL 语句的支持 echo=True
:
#### DELETING
#### COMMITING
2016-07-07 16:55:28,893 INFO sqlalchemy.engine.base.Engine SELECT persons.id AS persons_id, persons.name AS persons_name
FROM persons, car_ownerships
WHERE ? = car_ownerships.car_id AND persons.id = car_ownerships.person_id
2016-07-07 16:55:28,894 INFO sqlalchemy.engine.base.Engine (1,)
2016-07-07 16:55:28,895 INFO sqlalchemy.engine.base.Engine SELECT car_ownerships.id AS car_ownerships_id, car_ownerships.type AS car_ownerships_type, car_ownerships.car_id AS car_ownerships_car_id, car_ownerships.person_id AS car_ownerships_person_id
FROM car_ownerships
WHERE ? = car_ownerships.car_id
2016-07-07 16:55:28,896 INFO sqlalchemy.engine.base.Engine (1,)
2016-07-07 16:55:28,898 INFO sqlalchemy.engine.base.Engine DELETE FROM car_ownerships WHERE car_ownerships.car_id = ? AND car_ownerships.person_id = ?
2016-07-07 16:55:28,898 INFO sqlalchemy.engine.base.Engine ((1, 1), (1, 2))
2016-07-07 16:55:28,900 INFO sqlalchemy.engine.base.Engine UPDATE car_ownerships SET car_id=? WHERE car_ownerships.id = ?
2016-07-07 16:55:28,900 INFO sqlalchemy.engine.base.Engine ((None, 1), (None, 2))
2016-07-07 16:55:28,901 INFO sqlalchemy.engine.base.Engine ROLLBACK
编辑
使用association_proxy
我们可以使用关联代理来尝试实现 'through' 关系。
然而,为了直接.append()
一个依赖对象,我们需要为关联对象创建一个构造函数。此构造函数必须 'hacked' 才能成为双向的,因此我们可以同时使用两个赋值:
my_car.people.append(Person(name='my_son'))
my_husband.cars.append(Car(name='new_shiny_car'))
生成的(经过中间测试的)代码如下,但我对它不太满意(因为这个 hacky 构造函数还有什么会破坏?)。
编辑: 下面的 RazerM 的回答中介绍了使用关联代理的方法。 association_proxy()
有一个 creator 参数,它减轻了我最终在下面使用的巨大构造函数的需要。
class Person(Base):
__tablename__ = 'persons'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
cars = association_proxy('car_ownerships', 'car')
def __repr__(self):
return '<Person {} [{}]>'.format(self.name, self.id)
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
people = association_proxy('car_ownerships', 'person')
def __repr__(self):
return '<Car {} [{}]>'.format(self.name, self.id)
class CarOwnership(Base):
__tablename__ = 'car_ownerships'
id = Column(Integer(), primary_key=True)
type = Column(String(255))
car_id = Column(Integer(), ForeignKey(Car.id))
car = relationship('Car', backref='car_ownerships')
person_id = Column(Integer(), ForeignKey(Person.id))
person = relationship('Person', backref='car_ownerships')
def __init__(self, car=None, person=None, type='secondary'):
if isinstance(car, Person):
car, person = person, car
self.car = car
self.person = person
self.type = type
def __repr__(self):
return 'Ownership [{}]: {} <<-{}->> {}'.format(self.id, self.car, self.type, self.person)
您使用的是 Association Object,因此您需要采取不同的方式。
我改变了这里的关系,仔细看它们,因为一开始你有点难以理解(至少对我来说是这样!)。
我使用了 back_populates
,因为在这种情况下它比 backref
更清晰。多对多关系的双方都必须直接引用 CarOwnership
,因为这是您要使用的对象。这也是您的示例显示的内容;您需要使用它,以便您可以设置 type
.
class Person(Base):
__tablename__ = 'persons'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
cars = relationship('CarOwnership', back_populates='person')
def __repr__(self):
return '<Person {} [{}]>'.format(self.name, self.id)
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
people = relationship('CarOwnership', back_populates='car')
def __repr__(self):
return '<Car {} [{}]>'.format(self.name, self.id)
class CarOwnership(Base):
__tablename__ = 'car_ownerships'
id = Column(Integer(), primary_key=True)
type = Column(String(255))
car_id = Column(Integer(), ForeignKey(Car.id))
person_id = Column(Integer(), ForeignKey(Person.id))
car = relationship('Car', back_populates='people')
person = relationship('Person', back_populates='cars')
def __repr__(self):
return 'Ownership [{}]: {} <<-{}->> {}'.format(self.id, self.car, self.type, self.person)
请注意,删除任一侧后,car_ownerships
行不会被删除,它只会将外键设置为 NULL。如果您想设置自动删除,我可以在我的答案中添加更多内容。
编辑:直接访问Car
和Person
对象的集合,需要使用association_proxy
,类 然后改成这样:
from sqlalchemy.ext.associationproxy import association_proxy
class Person(Base):
__tablename__ = 'persons'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
cars = association_proxy(
'cars_association', 'car', creator=lambda c: CarOwnership(car=c))
def __repr__(self):
return '<Person {} [{}]>'.format(self.name, self.id)
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
people = association_proxy(
'people_association', 'person', creator=lambda p: CarOwnership(person=p))
def __repr__(self):
return '<Car {} [{}]>'.format(self.name, self.id)
class CarOwnership(Base):
__tablename__ = 'car_ownerships'
id = Column(Integer(), primary_key=True)
type = Column(String(255), default='secondary')
car_id = Column(Integer(), ForeignKey(Car.id))
person_id = Column(Integer(), ForeignKey(Person.id))
car = relationship('Car', backref='people_association')
person = relationship('Person', backref='cars_association')
def __repr__(self):
return 'Ownership [{}]: {} <<-{}->> {}'.format(self.id, self.car, self.type, self.person)
编辑: 在您的编辑中,您在将其转换为使用 backref
时犯了一个错误。您的汽车和人的关联代理不能同时使用 'car_ownerships' 关系,这就是为什么我有一个名为 'people_association' 和一个名为 'cars_association'.
您拥有的 'car_ownerships' 关系与关联 table 被称为 'car_ownerships' 这一事实无关,因此我以不同的方式命名它们。
我修改了上面的代码块。要允许附加工作,您需要将创建者添加到关联代理。我已将 back_populates
更改为 backref
,并将默认 type
添加到 Column
对象而不是构造函数。
最干净的解决方案如下,不涉及关联代理。它是完全成熟的多对多关系所缺少的秘诀。
在这里,我们编辑从依赖对象 Car 和 Person 到关联对象 的直接关系CarOwnership,为了防止这些关系在关联对象被删除后发出UPDATE
。为此,我们使用 passive_deletes='all'
标志。
产生的互动是:
- 能够从依赖对象查询和设置关联对象
# Changing Ownership type:
my_car.car_ownerships[0].type = 'primary'
# Creating an ownership between a car and a person directly:
CarOwnership(car=my_car, person=my_husband, type='primary')
直接访问和编辑依赖对象的能力:
# Get all cars from a person: [print(c) for c in my_husband.cars] # Update the name of one of my cars: me.cars[0].name = me.cars[0].name + ' Cabriolet'
创建或删除依赖对象时自动创建和删除关联对象
# Create a new owner and assign it to a car: my_car.people.append(Person('my_husband')) session.add(my_car) session.commit() # Creates the necessary CarOwnership # Delete a car: session.delete(my_car) session.commit() # Deletes all the related CarOwnership objects
代码
class Person(Base):
__tablename__ = 'persons'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
cars = relationship('Car', secondary='car_ownerships', backref='people')
def __repr__(self):
return '<Person {} [{}]>'.format(self.name, self.id)
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer(), primary_key=True)
name = Column(String(255))
def __repr__(self):
return '<Car {} [{}]>'.format(self.name, self.id)
class CarOwnership(Base):
__tablename__ = 'car_ownerships'
id = Column(Integer(), primary_key=True)
type = Column(String(255))
car_id = Column(Integer(), ForeignKey(Car.id))
car = relationship('Car', backref=backref('car_ownerships', passive_deletes='all'))
person_id = Column(Integer(), ForeignKey(Person.id))
person = relationship('Person', backref=backref('car_ownerships', passive_deletes='all'))
def __repr__(self):
return 'Ownership [{}]: {} <<-{}->> {}'.format(self.id, self.car, self.type, self.person)