区分父模型的Django继承的孩子 [英] Distinguishing parent model's children with Django inheritance

问题描述

基本上我有一个名为程序的Base类。然后我有更多具体的程序模型类型使用Program作为基类。对于99％的需求，我不在乎程序是否是特定子类型之一。当然，有1％的时间，我想知道是否是其中一个孩子。

问题是，如果我说，一个SwimProgram模型和使用Program作为其基础的CampProgram模型，发现它们是没有一堆try / except块的问题。我想要的是如下：

  program = models.Program.objects.get（id = 15）
如果program.swimprogram：
 ## do stuff 
 elif program.campprogram：
 ## do stuff 
 else：
 ##做其他的东西
  

当然这会抛出DoesNotExist异常。我可以使用更糟糕的try / excepts，或者我可以让Program有一个'type'字段，孩子们在保存时设置。两者都是可行的，但如果任何人有更好的方法，我很好奇。

解决方案

你尝试过hasattr（）吗？如下所示：

  if hasattr（program，'swimprogram'）：
＃... 
 elif hasattr（program，'campprogram'）：
＃... 
  

如果你是不确定这种方法，首先尝试在一个简单的测试应用程序。这是两个简单的模型，应该显示它是否适用于您和您使用的django版本（在django-1.1.1中测试）。

  class Archive（models.Model）：
 pub_date = models.DateField（）
 
 def __unicode __（self）：
 return存档：％s ％self.pub_date 
 
 class ArchiveB（Archive）：
 def __unicode __（self）：
 returnArchiveB：％s％self.pub_date 
  

然后在shell中给它一个旋转：

 > a_id = Archive.objects.create（pub_date =2010-10-10）。id 
> b_id = ArchiveB.objects.create（pub_date =2011-11-11）。id 
> a = Archive.objects.get（id = a_id）
> b = Archive.objects.get（id = b_id）
> （a，b）＃他们看起来像归档对象
（<存档：存档：2010-10-10>，<存档：存档：2011-11-11>）
> hasattr（a，'archiveb'）
 False 
> hasattr（b，'archiveb'）＃但只有一个可以访问ArchiveB 
 True 
  




Basically I have a Base class called "Program". I then have more specific program model types that use Program as a base class. For 99% of my needs, I don't care whether or not a Program is one of the specific child types. Of course there's that 1% of the time that I do want to know if it's one of the children.

The problem is that if I have let's say, a SwimProgram model and a CampProgram model using Program as their base, that it's problematic to find out what they are without a bunch of try/except blocks. What I want is something like the following:

program = models.Program.objects.get(id=15)
if program.swimprogram:
    ## do stuff
elif program.campprogram:
    ## do stuff
else:
    ## do other stuff

Of course this throws DoesNotExist exceptions. I could either use try/excepts which are uglier, or I could have Program have a 'type' field that the children set on save. Both are doable, but I'm curious if anyone has any better methods.

解决方案

Have you tried hasattr()? Something like this:

if hasattr(program, 'swimprogram'):
    # ...
elif hasattr(program, 'campprogram'):
    # ...

If you are unsure about this approach, try it out in a simple test app first. Here are two simple models that should show if it will work for you and the version of django that you are using (tested in django-1.1.1).

class Archive(models.Model):
    pub_date = models.DateField()

    def __unicode__(self):
        return "Archive: %s" % self.pub_date

class ArchiveB(Archive):
    def __unicode__(self):
        return "ArchiveB: %s" % self.pub_date

And then giving it a spin in the shell:

> a_id = Archive.objects.create(pub_date="2010-10-10").id
> b_id = ArchiveB.objects.create(pub_date="2011-11-11").id
> a = Archive.objects.get(id=a_id)
> b = Archive.objects.get(id=b_id)
> (a, b) # they both look like archive objects
(<Archive: Archive: 2010-10-10>, <Archive: Archive: 2011-11-11>)
> hasattr(a, 'archiveb')
False
> hasattr(b, 'archiveb') # but only one has access to an ArchiveB
True

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