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I am currently implementing a django app, for this I try to use a syntax that is consistent with Django's...

So here is what I am trying :

class Blablabla(Model):

    #this contains Blablabla's options
    class Meta:
        sort_key = lambda e: e

sort_key is a key function (for sorting purposes), but of course, it is understood as Meta's method (which is absolutely not what I want)!!!

Any workaround to this, that would still allow me to use this syntax ?

EDIT : Just an important precision ... the code I wrote is supposed to be written by somebody that uses the library ! That's why I don't want any dirty trick. And YES in Django it is really used just for options... of course Meta IS a class, but I say "it is not seen as a class", because it is not used as a class : you don't instantiate it, you don't put class methods, only class attributes... The Model has a metaclass that extracts everything from this Meta and handles all the options declared... But that's all ! It IS just a placeholder for options.

But OK that's True I never saw an option that is a function in Django... So I'll follow Ned an declare this sorting function as a method of Model that has to be overriden ...

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4 回答 4

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In general,

class Meta(object):
    sort_key= staticmethod(lambda e: e)

I've no idea if whatever magic Django does to transplant ‘meta’ members copes OK with decorated methods like this, but I don't see any inherent reason why not.

于 2010-09-16T23:36:41.127 回答
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你为什么要sort_key投入Meta？ Meta用于 Django 选项，它不是放置您自己的方法的地方。模型可以在其上定义方法。我想你想要一些简单的东西：

class Blablabla(Model):

    def sort_key(self, e):
        return e
于 2010-09-17T02:08:47.893 回答
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OP 在评论中写道“'Meta' 不应该被视为一个类”。如果是这样的话，也就是说，如果 Django 在Meta确实不是一个类（一个非常大的“如果”）时可以生存，那么就有可能满足 OP 避免使用最简单的解决方案的真正奇怪的愿望（只是stqticfunction围绕有lambda问题的解决方案） .

本质上，这需要编写一个（非常奇怪的）元类来生成一个对象，其中属性查找绕过类对描述符对象的正常使用（每个函数都是一个描述符对象：也就是说，它有一个__get__Python 通常在查找时使用的方法类或其实例上的属性）。

这种荒谬的旋转的一般想法是......：

class MetaNotAClass(type):
   def __new__(mcl, clasname, bases, clasdict):
     if bases:
         usedict = {}
     else:
         usedict = clasdict
     usedict['__foo'] = clasdict
     return type.__new__(mcl, clasname, bases, usedict)
   def __getattr__(cls, atname):
      try: return getattr(cls, '__foo')[atname]
      except KeyError: raise AttributeError, atname

class NotAClass:
  __metaclass__ = MetaNotAClass


class Bah(NotAClass):
  def f(): return 'weird!'

print Bah.f()

当然，任何期望 Bah成为类的东西都会破坏（但是，你确实说它“不应该被视为一个类”，所以这基本上就是你所要求的：破坏任何认为它是的代码“被视为一个阶级”！-)。

于 2010-09-17T00:18:17.903 回答
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你不能只创建一个简单的模块（meta？）并添加sort_key到它吗？然后继续前进，include无论您需要它...

于 2010-09-17T07:47:44.857 回答