我在django中遇到多表继承问题。
让我们以银行账户为例。
class account(models.Model):
name = models……
class accounttypeA(account):
balance = models.float…..
def addToBalance(self, value):
self.balance += value
class accounttypeB(account):
balance = models.int…. # NOTE this
def addToBalance(self, value):
value = do_some_thing_with_value(value) # NOTE this
self.balance += value
现在,我想为帐户类型添加一个值,但我拥有的只是一个帐户对象,例如acc = account.object.get(pk = 29)。那么,谁是acc的孩子?
Django会自动在accounttypeA和accounttypeB中创建account_ptr_id字段。所以,我的解决方案是:
child_class_list = ['accounttypeA', 'accounttypeB']
for cl in child_class_list:
try:
exec(“child = ” + str(cl) + “.objects.select_for_update().get(account_ptr_id=” + str(acc.id) + “)”)
logger.debug(“Child found and ready to use.”)
return child
except ObjectDoesNotExist:
logger.debug(“Object does not exist, moving on…”)
在这一点上,也许这是一个绘图板问题! :)
我希望我的例子清楚。感谢
答案 0 :(得分:8)
据我所知,没有Django内置方法可以做到这一点。
但是,给定acc=account.object.get(pk=29)
,您可以使用:
try:
typeA = acc.accounttypeA
# acc is typeA
except accounttypeA.DoesNotExist:
# acc should be typeB if account only has typeA and typeB subclasses
try:
typeB = acc.accounttypeB
# acc is typeB
except accounttypeB.DoesNotExist:
# acc should be typeA if account only has typeA and typeB subclasses
答案 1 :(得分:4)
我的解决方案基于this
class account(models.Model):
name = models……
def cast(self):
"""
This method is quite handy, it converts "self" into its correct child class. For example:
.. code-block:: python
class Fruit(models.Model):
name = models.CharField()
class Apple(Fruit):
pass
fruit = Fruit.objects.get(name='Granny Smith')
apple = fruit.cast()
:return self: A casted child class of self
"""
for name in dir(self):
try:
attr = getattr(self, name)
if isinstance(attr, self.__class__) and type(attr) != type(self):
return attr
except:
pass
@staticmethod
def allPossibleAccountTypes():
#this returns a list of all the subclasses of account (i.e. accounttypeA, accounttypeB etc)
return [str(subClass).split('.')[-1][:-2] for subClass in account.__subclasses__()]
def accountType(self):
try:
if type(self.cast()) == NoneType:
#it is a child
return self.__class__.__name__
else:
#it is a parent, i.e. an account
return str(type(self.cast())).split('.')[-1][:-2]
except:
logger.exception()
accountType.short_description = "Account type"
class accounttypeA(account):
balance = models.float…..
def addToBalance(self, value):
self.balance += value
class accounttypeB(account):
balance = models.int…. # NOTE this
答案 2 :(得分:3)
Django在课程account
中添加了两个字段:accounttypea
和accounttypeb
。如果你有pk = 42的accounttypeB
对象,你可以从父母那样访问:
account.objects.get(pk=42).accounttypeb
>>> <accounttypeB instance>
您可以为每个孩子的实际子类型添加CharField到父模型,然后使用getattr
,如果有很多子模型(它可能比很多try .. except xxx.DoesNotExist
更好块)。
class account(models.Model):
name = models……
cls = CharField(...)
def ext(self):
return getattr(self, self.cls.lower())
class accounttypeA(account):
balance = models.float…..
def addToBalance(self, value):
self.balance += value
class accounttypeB(account):
balance = models.int…. # NOTE this
def addToBalance(self, value):
value = do_some_thing_with_value(value) # NOTE this
self.balance += value
# example
accounttypeB.objects.create(balance=10, name='Vincent Law', cls="accounttypeB")
accounttypeA.objects.create(balance=9.5, name='Re-l Mayer', cls="accounttypeA")
for obj in account.objects.all():
obj.ext().addToBalance(1.0)
print(obj.name, obj.cls)
但你必须使用accounttypeA.objects.create(...)
和accounttypeB.objects.create(...)
创建模型 - 否则这个技巧将无效。
(https://docs.djangoproject.com/en/1.5/topics/db/models/#multi-table-inheritance)
答案 3 :(得分:0)
您可以使用hasattr()
这样的方法:
if hasattr(account, 'accounttypea'):
account.accounttypea.<somefield> = <some value>
do something here....
elif hasattr(account, 'accounttypeb'):
account.accounttypeb.<somefield> = <some value>
do something here...
不是很干但是可以用。 :)