使用父构造函数在SQLAlchemy会话中添加子类

Question

我有一个类继承方案，布局在http://docs.sqlalchemy.org/en/latest/orm/inheritance.html#joined-table-inheritance

from sqlalchemy import Column, Integer, String, ForeignKey
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

class Parent(Base):
    __tablename__ = 'parent'
    id = Column(Integer, primary_key=True)
    type = Column(String)
    __mapper_args__ = {'polymorphic_on': type}

class Child(Parent):
    __tablename__ = 'child'
    id = Column(Integer, ForeignKey('parent.id'), primary_key=True)
    __mapper_args__ = {'polymorphic_identity': 'child'}

我想能够使用Parent(如Parent(type='child'))的构造函数创建Child的实例，但它不起作用。当我启动ippython时…

In [1]: from stackoverflow.question import Parent, Child
In [2]: from sqlalchemy import create_engine
In [3]: from sqlalchemy.orm import sessionmaker
In [4]: session = sessionmaker(bind=create_engine(...), autocommit=True)()
In [5]: with session.begin():
    p = Parent(type='child')
    session.add(p)
   ...:     
/.../lib/python3.4/site-packages/sqlalchemy/orm/persistence.py:155: SAWarning: Flushing object <Parent at 0x7fe498378e10> with incompatible polymorphic identity 'child'; the object may not refresh and/or load correctly
  mapper._validate_polymorphic_identity(mapper, state, dict_)
In [6]: session.query(Parent).all()
Out[6]: [<stackoverflow.question.Parent at 0x7fe498378e10>]
In [7]: session.query(Child).all()
Out[7]: []

这可能吗?这是个好主意吗?

Answer 1

这绝对不是一个好主意。不需要使用构造函数来做一些修改，您可以只使用一个单独的辅助函数(工厂):

# create this manually
OBJ_TYPE_MAP = {
    # @note: using both None and 'parent', but should settle on one
    None: Parent, 'parent': Parent,
    'child': Child,
}
# ... or even automatically from the mappings:
OBJ_TYPE_MAP = {
    x.polymorphic_identity: x.class_
    for x in Parent.__mapper__.self_and_descendants
}
print(OBJ_TYPE_MAP)

def createNewObject(type_name, **kwargs):
    typ = OBJ_TYPE_MAP.get(type_name)
    assert typ, "Unknown type: {}".format(type_name)
    return typ(**kwargs)

a_parent = createNewObject(None, p_field1='parent_name1')
a_child = createNewObject(
    'child', p_field1='child_name1', c_field2='child_desc')
session.add_all([a_child, a_parent])

另一个注意事项:对于Parent，我将为{'polymorphic_identity': 'parent'}定义一个值。它比None更简洁。

EDIT-1: using Constructor

不是我推荐它，或者我真的知道我在这里做什么，但是如果您将__new__添加到Parent类:

def __new__(cls, *args, **kwargs):
    typ = kwargs.get('type')  # or .pop(...)
    if typ and not kwargs.pop('_my_hack', None):
        # print("Special handling for {}...".format(typ))
        if typ == 'parent':
            # here we can *properly* call the next in line
            return super(Parent, cls).__new__(cls, *args, **kwargs)
        elif typ == 'child':
            # @note: need this to avoid endless recursion
            kwargs["_my_hack"] = True
            # here we need to cheat somewhat
            return Child.__new__(Child, *args, **kwargs)
        else:
            raise Exception("nono")
    else:
        x = super(Parent, cls).__new__(cls, *args, **kwargs)
    return x

您将能够使用旧的方式(当没有type=xxx传递给__init__时)，或者通过提供参数来完成您的要求:

old_parent = Parent(field1=xxx, ...)
old_child = Child(field1=xxx, ...)
new_child = Parent(type='child', field1=xxx, ...)

同样，我不确定所有的含义，特别是因为sqlalchemy还覆盖了创建例程并使用了自己的元类。

Answer 2

问题是，当使用sqlalchemy声明性映射时，会为每个类生成一个映射器。

你要做的是让Parent的一个实例表现得像Child的一个实例，这是你做不到的，至少在不诉诸黑客的情况下。

从这个事实来看(你必须经过重重考验)这不是个好主意。也许你根本不需要继承?

编辑

如果你不想有条件逻辑或查找，你必须根据用户输入选择一个类，你可以这样做

cls = getattr(module_containing_the_classes, "<user_input>") 
cls(**kw)