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这些是什么,我有相关的位:如何部分覆盖/扩展声明类的构造函数?
import sqlalchemy as db
import sqlalchemy.ext.declarative
Base = db.ext.declarative.declarative_base()
class Product(Base):
__tablename__ = 'product'
id = db.Column(db.Integer, primary_key=True)
class Bin(Base):
__tablename__ = 'bin'
id = db.Column(db.Integer, primary_key=True)
product_id = db.Column(db.Integer, db.ForeignKey('product.id'), nullable=False)
product = db.orm.relationship('Product')
class PurchaseItem(Base):
__tablename__ = 'purchase_item'
id = db.Column(db.Integer, primary_key=True)
bin_id = db.Column(db.Integer, db.ForeignKey('bin.id'), nullable=False)
bin = db.orm.relationship('Bin')
我想什么是自动具有PurchaseItem
构造函数构造,如果它通过了Product
使用Bin
对象。我通常做的事:
def __init__(self, product=None, **kwargs):
if product is not None:
kwargs['bin'] = Bin(product=product)
super(PurchaseItem, self).__init__(self, **kwargs)
,但我得到这个错误:
>>> p = Product()
>>> pi = PurchaseItem(product=p)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<string>", line 4, in __init__
File "/Users/andrea/src/ifs/src/venv/lib/python2.7/site-packages/sqlalchemy/orm/state.py", line 306, in _initialize_instance
manager.dispatch.init_failure(self, args, kwargs)
File "/Users/andrea/src/ifs/src/venv/lib/python2.7/site-packages/sqlalchemy/util/langhelpers.py", line 60, in __exit__
compat.reraise(exc_type, exc_value, exc_tb)
File "/Users/andrea/src/ifs/src/venv/lib/python2.7/site-packages/sqlalchemy/orm/state.py", line 303, in _initialize_instance
return manager.original_init(*mixed[1:], **kwargs)
File "<stdin>", line 11, in __init__
TypeError: _declarative_constructor() takes exactly 1 argument (3 given)
大概是因为Base
是一个元类,并动态生成的子类构造函数。
我能够得到我想要的东西通过存储旧的构造,创造一个新的,然后调用旧:
_old_init = PurchaseItem.__init__
def _new_init(self, product=None, init=_old_init, **kwargs):
if product is not None:
kwargs['bin'] = Bin(product=product)
init(self, **kwargs)
PurchaseItem.__init__ = _new_init
有没有办法在PurchaseItem
类定义要做到这一点?否则,是否有一种方法不涉及临时变量ala emacs的defadvice
?