Get urls in views.py - python django

Question

my urls.py

url(r'^ACB/(?P<slug>.*)/$', views.ACB.as_view(), name='ACB'),

I have requirement where I need to send this url for couple of random slug as part of a response for an API

I do not want the url of request but url of different view

That is not very clear. Do you want the url of the request in your `ACB` view ? — Louis Barranqueiro, Dec 29 '15 at 14:09

Railslide · Accepted Answer · 2015-12-29T14:43:15.433

3

Use reverse

from django.core.urlresolvers import reverse

my_url = reverse('ACB')

This would reverse the ACB url without any argument. If slug is present, you can pass it in args:

my_url = reverse('ACB', args=(slug,))

edited Dec 29 '15 at 14:43

answered Dec 29 '15 at 14:11

Railslide

1

I would just update this with `my_url = reverse('ACB', args=(slug, ))` if present, as slug is an optional argument. – karthikr Dec 29 '15 at 14:15
It's amazing how you are able to get what OP wants. I still have trouble understanding his question. – Shang Wang Dec 29 '15 at 14:59

score 0 · Answer 2 · answered Jul 12 '21 at 21:26

Update 2021 (Django 2.0+)

In Django 2.0, the module django.core.urlresolvers was moved to django.urls.

New solution to question:

from django.urls import reverse

my_url = reverse('ACB')
my_url = reverse('ACB', args=(slug,))

For further information, see:

2 Answers2