Define searchFilter URL in django rest framework

Question

Currently I am working on a shop management project. The project is building with django, django-rest-framework and react. I want to filter the models data. So I have installed django-filter. I need to define the url like "http://localhost:8000/api/search-products/?category=oil&product=mustard oil". How can I do this?

My product search views is-

class SearchProduct(generics.ListAPIView):
    serializer_class = ProductSerializer
    filter_backends = [filters.SearchFilter]
    search_fields = ["category", "name"]

My url is-

path("api/search-product/?<str:category>&<str:name>", SearchProduct.as_view())

Its showing page not found error.

How can I define the appropriate url path?

Query params should not be part of the path url. They will still be covered though, and it will be handled by django-filter for you — Brian Destura
– Brian Destura, Commented Jul 5, 2021 at 5:10

Mojtaba Arezoomand · Accepted Answer · 2021-07-05 05:05:25Z

1

you shouldn't define it as path parameters:

path("api/search-product/", SearchProduct.as_view())

and you can call endpoint like this:

.../api/search-product/?search=any_thing_you_want

edited Jul 5, 2021 at 5:05

answered Jul 5, 2021 at 4:54

Mojtaba Arezoomand

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