Question

当您使用defer(...)查询命令时，Django会返回与原始模型不同的类。如何在使用延迟字段时动态获取模型的名称？

在代码中：

obj_nodefer = model_class.objects.filter(title="foo")[0]
model_name = str(type(obj_nodefer)) # Works just fine

obj_defer = model_class.objects.filter(title="foo").defer("content")[0]
model_name = str(type(obj_defer)) # Does't return the right name because of defer above.

如何从obj_defer获取模型的名称？

Answer 1

在延迟类上，您还可以使用以下方式获取原始类而不操纵__class__.__name__字符串：

obj._meta.concrete_model
'product.models.ProductModelName'

obj._meta.concrete_model._meta.app_label
'product'

obj._meta.concrete_model._meta.model_name
'productmodelname'

Answer 2

要获得非延迟模型，请执行

t = type(obj)
if t._deferred:
  t = t.__base__

Answer 3

您可以访问obj.__class__

  >>> print obj.__class__
  <class 'product.models.Product_Deferred_name'>
  >>> print obj.__class__.__name__
  Product_Deferred_name
  >>> print obj.__class__.__name__.split("_")[0]
  Product

如果您的型号名称包含_，

  >>> print obj.__class__.__name__.replace("_Deferred_name","")
  Product

Answer 4

在django 1.4中，延迟类的构造如下：

def deferred_class_factory(model, attrs):
    """
    Returns a class object that is a copy of "model" with the specified "attrs"
    being replaced with DeferredAttribute objects. The "pk_value" ties the
    deferred attributes to a particular instance of the model.
    """
    class Meta:
        proxy = True
        app_label = model._meta.app_label

    # The app_cache wants a unique name for each model, otherwise the new class
    # won't be created (we get an old one back). Therefore, we generate the
    # name using the passed in attrs. It's OK to reuse an existing class
    # object if the attrs are identical.
    name = "%s_Deferred_%s" % (model.__name__, '_'.join(sorted(list(attrs))))
    name = util.truncate_name(name, 80, 32)

    overrides = dict([(attr, DeferredAttribute(attr, model))
            for attr in attrs])
    overrides["Meta"] = Meta
    overrides["__module__"] = model.__module__
    overrides["_deferred"] = True
    return type(name, (model,), overrides)

所以，从理论上讲，你应该通过

进入真正的班级

type(type(deferred))

Django：使用延迟查询时获取模型类型

4 个答案: