How to implement search view for list view?

Django (web framework): How can I implement a urlpattern that passes a list of strings to the view?

• For example: I have a url like this: http://mysite/StringA/StringB/StringN And I want to always process this with the same view, but I would like the view to receive a list ['StringA', 'StringB', 'StringN'] for processing. Is there an easy way to setup http://urls.py to pass this information along, or would it be easier to just setup a regex to capture any URL and have the view parse the URL afterwards?

• Answer:

  I have been experimenting and there are several ways to do this, the summary is that you just need to match all characters and front slashes. If you want your URLs to include numbers or other characters, you need to include that too. In your http://urls.py (preferably after your other routes, since this stands to intercept a lot): 1(r'^/([A-Za-z/]+)','home.views.blarg'), And assuming you have a controller called home, in http://views.py you need a method like this: 1def blarg(request, arguments): Here, arguments will contain a string of the fully called path, excluding the base of home. As per the question, you can split it into a list like so: 1arglist = arguments.split('/')