Question

我的问题是关于SQLAlchemy，但我在解释它时遇到了麻烦所以我想我用一个简单的例子来解释它我想要实现的目标：

parent = Table('parent', metadata,
    Column('parent_id', Integer, primary_key=True),
    Column('name', Unicode),
)
parent_child = Table('parent_child', metadata,
    Column('parent_id', Integer, primary_key=True),
    Column('child_id', Integer, primary_key=True),
    Column('number', Integer),
    ForeignKeyConstraint(['parent_id'], ['parent.parent_id']),
    ForeignKeyConstraint(['child_id'], ['child.child_id']),
)
child = Table('child', metadata,
    Column('child_id', Integer, primary_key=True),
    Column('name', Unicode),
)
class Parent(object):
    pass
class ParentChild(object):
    pass
class Child(object):
    pass

>>> p = Parent(name=u'A')
>>> print p.children 
{}
>>> p.children[0] = Child(name=u'Child A')
>>> p.children[10] = Child(name=u'Child B')
>>> p.children[10] = Child(name=u'Child C')

此代码将在parent_child表中创建3行，第一行的列号为0，第二行和第三行的列号为10。

>>> print p.children 
{0: [<Child A>], 10: [<Child B>, <Child C>]}
>>> print p.children[10][0]
<Child B>

（我在示例中遗漏了所有SQLAlchemy会话/引擎代码，以使其尽可能干净）

我尝试使用'

collection_class=attribute_mapped_collection('number')

关于Parent和ParentChild之间的关系，但它只给了我一个孩子的每个数字。不是带有列表的字典。任何帮助表示赞赏！

已更新！

感谢Denis Otkidach，我现在有了这段代码，但它仍然不起作用。

def _create_children(number, child):
    return ParentChild(parent=None, child=child, number=number)

class Parent(object):
    children = association_proxy('_children', 'child', creator=_create_children)

class MyMappedCollection(MappedCollection):
    def __init__(self):
        keyfunc = lambda attr_name: operator.attrgetter('number')
        MappedCollection.__init__(self, keyfunc=keyfunc)

    @collection.appender
    @collection.internally_instrumented
    def set(self, value, _sa_initiator=None):
        key = self.keyfunc(value)
        try:
            self.__getitem__(key).append(value)
        except KeyError:
            self.__setitem__(key, [value])

mapper(Parent, parent, properties={
     '_children': relation(ParentChild, collection_class=MyMappedCollection),
})

插入Child似乎可以正常工作

p.children[100] = Child(...)

但是当我尝试打印这样的孩子时：

print p.children

我收到此错误：

sqlalchemy.orm.exc.UnmappedInstanceError: Class '__builtin__.list' is not mapped

Answer 1

您必须定义自己的集合类。只有3种方法可以实现：appender，remover和converter。请参阅sqlalchemy.orm.collections.MappedCollection作为示例。

更新：根据您的要求，这是一个快速实施的方法：

from sqlalchemy import *
from sqlalchemy.orm import mapper, relation, sessionmaker
from sqlalchemy.orm.collections import collection

metadata = MetaData()

parent = Table('parent', metadata,
    Column('parent_id', Integer, primary_key=True),
    Column('name', Unicode),
)

child = Table('child', metadata,
    Column('child_id', Integer, primary_key=True),
    Column('name', Unicode),
)

parent_child = Table('parent_child', metadata,
    Column('parent_id', Integer, ForeignKey(parent.c.parent_id)),
    Column('child_id', Integer, ForeignKey(child.c.child_id)),
    Column('number', Integer),
    PrimaryKeyConstraint('parent_id', 'child_id'),
)

class ParentChild(object):
    def __init__(self, child, number):
        self.child = child
        self.number = number

class Parent(object): pass

class Child(object): pass


class MyMappedCollection(object):

    def __init__(self, data=None):
        self._data = data or {}

    @collection.appender
    def _append(self, parent_child):
        l = self._data.setdefault(parent_child.number, [])
        l.append(parent_child)

    def __setitem__(self, number, child):
        self._append(ParentChild(number=number, child=child))

    def __getitem__(self, number):
        return tuple(pc.child for pc in self._data[number])

    @collection.remover
    def _remove(self, parent_child):
        self._data[parent_child.number].remove(parent_child)

    @collection.iterator
    def _iterator(self):
        for pcs in self._data.itervalues():
            for pc in pcs:
                yield pc

    def __repr__(self):
        return '%s(%r)' % (type(self).__name__, self._data)


mapper(Parent, parent, properties={
     'children': relation(ParentChild, collection_class=MyMappedCollection),
})
mapper(Child, child)
mapper(ParentChild, parent_child, properties={
    'parent': relation(Parent),
    'child': relation(Child),
})

engine = create_engine('sqlite://')
db = sessionmaker(bind=engine)()
metadata.create_all(bind=engine)

p = Parent()
c1 = Child()
c2 = Child()
c3 = Child()
p.children[1] = c1
p.children[1] = c2
p.children[2] = c3

db.add(p)
db.commit()
p_id = p.parent_id
db.expunge_all()

p = db.query(Parent).get(p_id)
print p.children[1]
print p.children[2]

使用带有SQLAlchemy的中间人表作为列表的dict属性

已更新！

1 个答案: