Get slug name list by url path name - django

Question

I am reverse-engineering django urlpatterns. For my purpose, I need to dynamically know the list of slug names developers chose to use within url. (for weird decorator purposes outside of scope of this question)
Example:
path("<slug:membership_slug>/<slug:parent_slug>/data/", rzd.ReportingUnitDataView.as_view(), name="hierarchy.item.data"),
So, in the world, where everyone writes perfect code, my function whould take "hierarcy.item.data" and spit out ["membership_slug", "parent_slug"]
I looked at the reverse function, hoping it would return something useful without kwargs, but it does not. Have been googling like Neo, but the hope does not leave my heart... somewhere django should hold all that urlpatterns stuff! I still believe.
Perhaps you can at suggest how to get a hold of list of urls, even if in the raw format, that would already be a help. Thank you.

Answer 1

Search for book called Django 2x scopes this book hold every thing related to django best practices and of course the urls part.
This book will help you to understand more how django developers think.
Happy hacking 🙃

Answer 2

I think you're best guess is to write a Regex..
From what I've found in the source code they use _lazy_re_compile(r'<(?:(?P<converter>[^>:]+):)?(?P<parameter>[^>]+)>') which can be found in urls\resolvers.py in _route_to_regex
this is what I've come up with after butchering the source code
def geturlnamedarguments(route):
from django.utils.regex_helper import _lazy_re_compile
named_arguments = []
while True:
match = _lazy_re_compile(r'<(?:(?P<converter>[^>:]+):)?(?P<parameter>[^>]+)>').search(route)
if not match:
break
route = route[match.end():]
parameter = match['parameter']
named_arguments.append(parameter)
return named_arguments
You can also get the raw URL route as a string by doing
urlpatterns[index].pattern._route
You can find the URLPattern Object also in urls\resolvers.py if you want to play around with it.. I never found a place where they have the names in a raw format

Answer 3

Solved, It occurred to me, that I can just include urls.py and parse it out myself :)
from myapp import urls
class UrlPatterns:
#staticmethod
def _get_params(s):
items = []
is_exp = False
is_prefix = False
type = ''
name = ''
for n in range(len(s)):
if not is_exp:
if s[n] == '<':
is_exp = True
is_prefix = True
type = ''
elif is_prefix:
if s[n] == ':':
is_prefix = False
else:
type += s[n]
else:
if s[n] == '>':
is_exp = False
items.append({
'type': type,
'name': name
})
name = ''
else:
name += s[n]
return items
#staticmethod
def _find_pattern(url_name):
for url in urls.urlpatterns:
if url.name == url_name:
return str(url.pattern)
return None
#staticmethod
def get_required_params(url_name):
url_pattern = UrlPatterns._find_pattern(url_name)
if url_pattern:
return UrlPatterns._get_params(url_pattern)
return None