How to write the urls in urls.py in django?

Question

I am trying to write a class-based view which will take two inputs, say "city" and "bankname," and filter the data and send the results.

I am trying to write:

http://127.0.0.1:8000/api/bankcitydetail?city=mumbai&bankname=CENTRAL BANK OF INDIA. 

How would I write the URL's path in urls.py?

Answer 1

from django.urls import path
from . import views

urlpatterns = [
   path('', views.x)
]

x = Whatever your function is

Answer 2

I am not sure why you included the bankcitydetail in your url, you could just use:

http://127.0.0.1:8000/api/?city=mumbai&bankname=CENTRAL BANK OF INDIA

That would be something like this:

from django.urls import path
from . import views

urlpatterns = [
    path('', views.index, name='index'),
    path('api/', views.api, name='api'),
]

If you want to use the bankcitydetail to specify the resource in the API, you could use path converters like in the example in Django'sdocumentation .