URL pattern for query string parameters Django REST Framework

Question

This is my current url path is as http://localhost:8000/api/projects/abcml/2021 and in the urls.py page I am passing as path("api/projects/<str:project_handle>/<int:year>", functionname...) and in the view, I accept this parameters with self.kwargs.get method.

I want to pass url as this format http://localhost:8000/api/projects/abcml/?year=2021 What changes do I need to make in url pattern?

I tried this path("api/projects/<str:project_handle>/?year=<int:year> but did not seem correct also in the view page, instead of self.kwargs.get I changed it to self.request.query_params.get for year parameter. That did not work either. Error it throwing is Page not found (404).

willeM_ Van Onsem · Accepted Answer · 2022-06-22 22:09:52Z

3

The query string [wiki] is not part of the path, and therefore can not be matched.

You thus specify as path:

path('api/projects/<str:project_handle>/', functionname)

In the view, you can access the data with self.request.GET['year'] and this will return a string or a KeyError in cas the year was not provided in the query string.

answered Jun 22, 2022 at 22:09

willeM_ Van Onsem

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6 Comments

Hardik Patil Over a year ago

Its throwing 'argument of type 'Request' is not iterable'

willeM_ Van Onsem Over a year ago

@HardikPatil: likely the error is somewhere else, for example by iterating over a request object somehow...

Hardik Patil Over a year ago

How to check whether this query parameter year is empty or not in the view ?

willeM_ Van Onsem Over a year ago

@HardikPatil: if 'year' in self.request.GET.

Hardik Patil Over a year ago

Yeah also along with that, had to check 'if not' condition inside 'if in' condition. Hey thank you! Btw how did you add code block section in the comment?

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