The current path, book/1/2/3/4/5/6/, didn’t match any of these

Question

I've Created a view

urlpatterns = [
   path('book/<suffix>/', views.bookview, name='bookDetail')`
]

what i want

if someone hits url '127.0.0.1:8000/book/1/2/3/4/5/6/'
i do not want django to raise an error of The current path, book/1/2/3/4/5/6/, didn’t match any of these.
but instead it shows the same view of url book/<suffix>/ which is view.bookview.
and somehow pass, suffix after book/ which is 1/2/3/4/5/6/ as arg so that i can access it in my view.

See this https://stackoverflow.com/questions/33555195/django-urlpattern-with-infinite-number-of-parameters — Razenstein, Jun 03 '22 at 04:51
Does this answer your question? [How one can capture string that contain one or more forward slash in django urls](https://stackoverflow.com/questions/68251393/how-one-can-capture-string-that-contain-one-or-more-forward-slash-in-django-urls) — Abdul Aziz Barkat, Jun 03 '22 at 04:51

score 0 · Answer 1 · answered Jun 03 '22 at 05:06

0

from django.urls import re_path

urlpatterns = [
    re_path(r'^book/(?P<suffix>[\w/]+)$', views.bookview, name='bookDetail')
]

answered Jun 03 '22 at 05:06

Verma

score 0 · Answer 2 · answered Jun 03 '22 at 05:08

0

As there can be special characters in the suffix parameter, you need to encode the string. In python you can do it like this

from urllib.parse import quote_plus
suffix = quote_plus('/1/2/3/4/5/6/')

answered Jun 03 '22 at 05:08

salmanwahed

2 Answers2