从Django中的基本模型实例返回代理模型实例的正确方法？ [英] Right way to return proxy model instance from a base model instance in Django?

问题描述

说我有模型：

  class Animal（models.Model）：
 type = models.CharField max_length = 255）
 
 class Dog（Animal）：
 def make_sound（self）：
 printWoof！ 
 class Meta：
 proxy = True 
 
 class Cat（Animal）：
 def make_sound（self）：
 print喵！ 
 class Meta：
 proxy = True 
  

让我们说要做：

  animals = Animal.objects.all（）
动物动物：
 animal.make_sound ）
  

我想收回一系列Woofs和Meows。显然，我可以在原始模型中定义一个基于animal_type的一个make_sound，但是每当我添加一个新的动物类型（想象它们在不同的应用程序中）时，我必须进入并编辑make_sound函数。我宁愿只是定义代理模型，并让他们自己定义行为。从我可以看出来，没有办法返回混合的Cat或Dog实例，但是我想也许我可以在主类上定义一个返回一个猫或狗模型的get_proxy_model方法。

当然你可以这样做，并传递一些像主键，然后只是做Cat.objects.get（pk = pass_in_primary_key）。但这意味着对已经拥有的数据进行额外的查询，这似乎是多余的。有没有办法有效地将动物变成猫或狗的实例？什么是正确的方法来做我想要实现的？

解决方案

你可以使用描述的方法使Django模型多态href =http://groups.google.com/group/django-developers/msg/b567b75fe61d5044 =nofollow noreferrer> here 。该代码处于开发的早期阶段，我相信，但值得调查。

Say I have models:

class Animal(models.Model):
    type = models.CharField(max_length=255)

class Dog(Animal):
    def make_sound(self):
        print "Woof!"
    class Meta:
        proxy = True

class Cat(Animal):
    def make_sound(self):
        print "Meow!"
    class Meta:
        proxy = True

Let's say I want to do:

 animals = Animal.objects.all()
 for animal in animals:
     animal.make_sound()

I want to get back a series of Woofs and Meows. Clearly, I could just define a make_sound in the original model that forks based on animal_type, but then every time I add a new animal type (imagine they're in different apps), I'd have to go in and edit that make_sound function. I'd rather just define proxy models and have them define the behavior themselves. From what I can tell, there's no way of returning mixed Cat or Dog instances, but I figured maybe I could define a "get_proxy_model" method on the main class that returns a cat or a dog model.

Surely you could do this, and pass something like the primary key and then just do Cat.objects.get(pk = passed_in_primary_key). But that'd mean doing an extra query for data you already have which seems redundant. Is there any way to turn an animal into a cat or a dog instance in an efficient way? What's the right way to do what I want to achieve?

解决方案

You can perhaps make Django models polymorphic using the approach described here. That code is in early stages of development, I believe, but worth investigating.

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