Django-tastypie：如何以编程方式进行ModelResource？

Question

I have created a number of Django-tastypie resource definitions in resources.py. Most of them will be created by posting from javascript side (backbone-tastypie), but some of them I would like to be able to create right from python code/django views. The reason for this is that object creation logic should be kept in one place, as well as authorization params.

Is there some neat way to create TastyPie ModelResource inside Python code? (may be making a "post" with "Requests" module?).

Answer 1

如果我理解正确，那么您的用例将在好吃的食谱中描述： http ://django-tastypie.readthedocs.org/en/latest/cookbook.html#using-your-resource-in-regular-views