在Django中仅列出项目作为URL参数

Question

Id like to have url variable that would only accept items from list, but i dont know how to implement it in urls.

list = ['foo', 'bar', 'test', 'test2', 'random']

and urls.py, that i dont know how to implement list items:

url(r'^(?P<list_item>)/$', views.letnik, name="list_item"),

I would really appreciate it if you could help me out!

Answer 1

You can restrict this in views easily. In your view "letnik" can handle this by checking kwargs.
kwargs.get('list_item') will have the pattern matched.
You can check whether the parameter in url matches one in your list and if it does not you can either redirect to another url or throw 404 status.

One way to do this is by restricting in urls.py is as follows:

url(r'^(?P<list_item>foo|bar|test|test2|random)/$', views.letnik, name="list_item"),

But as the list expands it is better to handle it in views.