我试图在模板中迭代子模型实例的属性,具体来说,我只想访问childs属性。在运行时,我不知道它是什么具体的子类。使用django-model-utils,我能够返回子类而不是父类,这是一个开始,但是当我访问它的属性时,我会得到父类和子类的返回:
class Product(models.Model):
created_at = models.DateTimeField(default=timezone.now)
updated_at = models.DateTimeField(auto_now=True)
name = models.CharField(...)
objects = InheritanceManager()
def attrs(self):
for attr, value in self.__dict__.iteritems():
yield attr, value
class Vacuum(Product):
power = models.DecimalField(...)
class Toaster(Product):
weight = models.DecimalField(...)views.py
def product_detail(request, slug):
product = Product.objects.get_subclass(slug=slug)模板
{% for name, value in product.attrs %}
<td>{{ name }}</td>
<td>{{ value }}</td>
{% endfor %}发布于 2017-02-21 15:44:52
你能这样做吗?
def product_detail(request, slug):
product = Product.objects.get_subclass(slug=slug)
child_fields = [i for i in product.__class__.__dict__ if
not i.startswith("__") and not hasattr(Product, i)]
product_attrs = [(name, getattr(product,name)) for name in child_fields]
# pass product_attrs to your templatehttps://stackoverflow.com/questions/42371028
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