How to set optional parameter in a django url?

Question

I have write down this code in a url

path('list/<str:name>', GetList.as_view(), name = "getList" )

my view.py

class BlogList(View):
  def get(self,request, *args, **kwargs):

now i want to set name as optional parameter with list it show show all and with name i will implement query

ilyasbbu · Accepted Answer · 2022-11-02T06:51:29.137

2

There are multiple approaches to do this, One simple approach is to have multiple rules that matches your needs, all pointing to the same view.

urlpatterns = patterns('',
    url(r'^project_config/$', views.foo),
    url(r'^project_config/(?P<product>\w+)/$', views.foo),
    url(r'^project_config/(?P<product>\w+)/(?P<project_id>\w+)/$', views.foo),
)

Also keep in mind that in your view you'll also need to set a default for the optional URL parameter, or you'll get an error:

def foo(request, optional_parameter=''):
    # Your code goes here

For Django version > 2.0

Similar approach , but syntax

urlpatterns = [
    path('project_config/',views.foo,name='project_config'),
    path('project_config/<product>/',views.foo,name='project_config'),
    path('project_config/<product>/<project_id>/',views.foo,name='project_config'),
]

edited Nov 02 '22 at 06:51

answered Nov 02 '22 at 06:42

ilyasbbu

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please done copy and paste ans it not working and watch my code i am using class view ok – Jaskaran Singh Nov 02 '22 at 07:06
Copy pasting won't work, what I shown here is just example on how to implement – ilyasbbu Nov 02 '22 at 07:07
with class view ? – Jaskaran Singh Nov 02 '22 at 07:08
path('list/', BlogList.as_view(), name = "bloglist" ), path('list/', BlogList.as_view(), name = "bloglist" ), i have added 2 url – Jaskaran Singh Nov 02 '22 at 07:17

How to set optional parameter in a django url?

1 Answers1

For Django version > 2.0