使用多态模型进行SQLALchemy多对多模型关系配置
使用多态模型进行SQLALchemy多对多模型关系配置
提问于 2015-12-23 06:52:58
所以，有一些问题和答案涉及到这个问题，但我不能使它们与我试图实现的目标完全一致。
我有一套模型，它们是自引用的，并且是继承的。这是基本的设计。
class BaseUser(db.Model):
id = db.Column(db.Integer, primary_key=True, nullable=False)
org = db.Column(db.Boolean, default=False, nullable=False)
# Shared Fields
__mapper_args__ = {
'polymorphic_on': org,
}
class Customer(BaseUser):
# Customer Fields
__mapper_args__ = {
'polymorphic_identity': 0
}
class Organization(BaseUser):
# Organization Fields
__mapper_args__ = {
'polymorphic_identity': 1
}
class CustomerOrganization(db.Model):
user_id = db.Column(db.ForeignKey('customer.id', ondelete=CASCADE, onupdate=CASCADE), primary_key=True, nullable=False)
org_id = db.Column(db.ForeignKey('customer.id', ondelete=CASCADE, onupdate=CASCADE), primary_key=True, nullable=False)我尝试了几种不同的方法来创建每种类型的“组织”和“成员”关系。关于如何定义relationsihp()属性有什么建议吗？
回答 1
Stack Overflow用户
发布于 2016-01-17 04:11:47
这可以使用primaryjoin和secondaryjoin属性来完成。相关文档为here。
示例：
customer_organization = Table(
'base_user_customer_organization', ModelBase.metadata,
Column('user_id', Integer, ForeignKey('base_user.id')),
Column('org_id', Integer, ForeignKey('base_user.id'))
)
class BaseUser(ModelBase):
__tablename__ = 'base_user'
id = Column(Integer, primary_key=True, nullable=False)
org = Column(Boolean, default=False, nullable=False)
# Shared Fields
__mapper_args__ = {
'polymorphic_on': org,
}
customers = relationship(
"BaseUser",
backref=backref('organization', order_by=id),
secondary=customer_organization,
primaryjoin=id==customer_organization.c.org_id and org==True,
secondaryjoin=id==customer_organization.c.user_id and org==False
)
class CustomerUser(BaseUser):
# Customer Fields
__mapper_args__ = {
'polymorphic_identity': False
}
class OrganizationUser(BaseUser):
# Organization Fields
__mapper_args__ = {
'polymorphic_identity': True
}和测试：
sql = sqldb.get_session()
customer1 = sqldb.system.CustomerUser()
sql.add(customer1)
customer2 = sqldb.system.CustomerUser()
sql.add(customer2)
organization = sqldb.system.OrganizationUser()
organization.customers = [customer1, customer2]
sql.add(organization)
sql.commit()
# function prints all table data
print get_sql_table_data(sqldb.system.BaseUser)
print organization.customers
print customer1.organization
print customer2.organization输出：
[{'org': False, 'id': 1}, {'org': False, 'id': 2}, {'org': True, 'id': 3}]
[<CustomerUser(id=1, org=False)>, <CustomerUser(id=2, org=False)>]
[<OrganizationUser(id=3, org=True)>]
[<OrganizationUser(id=3, org=True)>]页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/34425777
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