如何在Django中将URL分解为模型和参数

Question

Basically, I want the reverse version of get_absolute_url method in Django model. 基本上，我想要Django模型中get_absolute_url方法的反向版本。

For example, if we have a model: 例如，如果我们有一个模型：

class MyModel(Model):
    ...

    get_absolute_url(self):
        return reverse('myview', args=(myparam=self.myfield,))

then in some view we should be able to do something like this: 那么从某种角度来说，我们应该能够执行以下操作：

url = 'http://www.example.com/path/to/myview/123'
model, params = decompose(url)
# now model="MyModel", params=('123',)

EDIT: 编辑：

If it's too difficult to achieve, how about re-reversing views only? 如果太难实现了，那么只重新反转视图呢？

url = 'http://www.example.com/path/to/myview/123'
view, params = decompose(url)
# now view="MyView", params=('123',)

Answer 1

from django.core import urlresolvers

match = urlresolvers.resolve(urlsplit(url).path)
view = match.func
params = match.args

Solution from django-users . django-users的解决方案。