与方炼金术关系的动态行为

Question

我有两个表，和一个关系一对多，定义后。

from sqlalchemy import Column, Integer, ForeignKey
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship

Base = declarative_base()


class Parent(Base):
    __tablename__ = 'parent'
    id = Column(Integer, primary_key=True)
    children = relationship("Child", backref="parent")


class Child(Base):
    __tablename__ = 'child'
    id = Column(Integer, primary_key=True)
    parent_id = Column(Integer, ForeignKey('parent.id'))

如果我试图访问Child.parent，就会得到一个错误。但是，如果初始化子实例，错误就会消失。

我猜初始化一个修改过的Child类，但是我不明白如何初始化。在不创建子实例的情况下，我如何能够访问Child.parent？

In [1]: Child.parent
---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-2-8b757eeb36c4> in <module>()
----> 1 Child.parent

AttributeError: type object 'Child' has no attribute 'parent'

In [2]: Child()
Out[2]: <etl.es_utils.test_to_rm.Child at 0x7f38caf42c50>

In [3]: Child.parent

Answer 1

我没有使用backref，而是定义了双方的关系，它解决了这个问题。

from sqlalchemy import Column, Integer, ForeignKey
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship

Base = declarative_base()


class Parent(Base):
    __tablename__ = 'parent'
    id = Column(Integer, primary_key=True)
    children = relationship("Child")


class Child(Base):
    __tablename__ = 'child'
    id = Column(Integer, primary_key=True)
    parent_id = Column(Integer, ForeignKey('parent.id'))
    parent = relationship("Parent")