CODEWITHSUNDEEP
pythondjangodjango-rest-framework
HomeAlone
HomeAlone

Reputation: 183

Django Rest Framework URL pattern with parameter

I have an Angular/DRF application.

What I want is to send an object to my backend to process it:

 var data = {'url': url, 'name': name}
 return this.http.get("http://127.0.0.1:8000/api/record/json=" + encodeURIComponent(JSON.stringify(data)))

Because my data as URL in it, I URI encode it to escape the "/"

Then in DRF this my urls.py:

path('record/<str:music_details>', record_views.Record.as_view(), name='record'),

And my views.py:

class Record(View):

    def get(self, request):
        json_string = request.query_params['json']
        data = json.loads(json_string)
        print(data)

I have nothing printing in my console, I assume my url path is wrong, how should I fix it?

Upvotes: 1

Views: 1890

Answers (1)

Lovepreet Singh
Lovepreet Singh

Reputation: 4840

Solution 1:

Send query parameter after ? like:

var data = {'url': url, 'name': name}
return this.http.get("http://127.0.0.1:8000/api/record?json=" + encodeURIComponent(JSON.stringify(data)))

Modify urls.py:

path('record', record_views.Record.as_view(), name='record'),

and fetch from query as:

json_string = request.GET.get('json')

Solution 2:

Send data with json= like:

var data = {'url': url, 'name': name}
return this.http.get("http://127.0.0.1:8000/api/record/" + encodeURIComponent(JSON.stringify(data)))

and modify get method to fetch data as:

class Record(View):

    def get(self, request, music_details):
        data = json.loads(music_details)
        print(data)

Upvotes: 3

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