Why is my Django rest framework api returning an empty list?

Question

Hello I am working on an Django Rest Framework api and one of the views is returning an empty list.

here's my view:

@api_view(['GET'])
def post_search(request):
    form = SearchForm()
    query = None
    results = []
    if request.method == 'GET':
        if 'query' in request.GET:
            form = SearchForm(request.GET)
            if form.is_valid():
                query = form.cleaned_data['query']
                results = Paper.objects.annotate(
                    search=SearchVector('abstract', 'title'),).filter(search=query)
        serializer = PaperSerializer(results, many=True)

        return Response(serializer.data)

here is the form:

class SearchForm(forms.Form):
    query = forms.CharField()

and here are my urls:

path('search/', views.post_search, name='post_search'),

So on the shell I ran:

Paper.objects.annotate(search=SearchVector('abstract', 'title'),).filter(search='Type')

and I got the results I wanted but when do this:

import requests
url = 'http://127.0.0.1:8000/api/search/?search=Type'
re = requests.get(url)
re.json # -> []
# or
import json
json.loads(re) # ->raise TypeError(f'the JSON object must be str, # bytes or bytearray, '
#TypeError: the JSON object must be str, bytes or bytearray, not Response

Any help will be appreciated; thanks

@sahasrara62 how do I check this? sorry for my ignorance but im new to web dev. — chez93
– chez93, Commented Jan 7, 2023 at 23:12
You can add import pdb; pdb,set_trace(), after if request.method == 'GET': and restart application and then hit the end point, in terminal run 'query' in request.GET, if it is False then is root cause of your error — sahasrara62
– sahasrara62, Commented Jan 7, 2023 at 23:15
'http://127.0.0.1:8000/api/search/?search=Type' That url does not contain a query arg, so if 'query' in request.GET will be false. — John Gordon
– John Gordon, Commented Jan 7, 2023 at 23:30

willeM_ Van Onsem · Accepted Answer · 2023-01-07 23:35:48Z

1

Your parameter is search not ~~query~~, so you check with if 'search' in request.GET. But you make things overcomplicated. You can work with:

class SearchForm(forms.Form):
    search = forms.CharField()


@api_view(['GET'])
def post_search(request):
    form = SearchForm(request.GET)
    if form.is_valid():
        results = Paper.objects.annotate(
            search=SearchVector('abstract', 'title')
        ).filter(search=form.cleaned_data['search'])
        serializer = PaperSerializer(results, many=True)
        return Response(serializer.data)
    return Response({'error': 'invalid search'}, status=400)

answered Jan 7, 2023 at 23:35

willeM_ Van Onsem

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5 Comments

chez93 Over a year ago

I get the error: Bad Request: /api/search/ [07/Jan/2023 23:43:08] "GET /api/search/?search=Type HTTP/1.1" 400 26

willeM_ Van Onsem Over a year ago

@chez93: looks like it never "fires" the view, so something is wrong with your urls.py probably. Did you define a view with the same name? Did you forget to include the urlpatterns somewhere in the "top level" urls.py?

chez93 Over a year ago

here is my top level url: path('api/', include('restapi.urls')),

dsenese Over a year ago

Try to hit the endpoint without the slash between the url and the parameter. Like this: http://127.0.0.1:8000/api/search?search=Type

chez93 Over a year ago

@dsenese I tried bout urls - with slash and without it -- and when I do re.json() I get {'error' : 'invalid search'}

chez93 · Accepted Answer · 2023-01-08 03:56:17Z

0

The problem was the form.

The form as I had was:

class SearchForm(forms.Form):
    query = forms.CharField()

the field here query didn't match with search in the url. So all I did was change the form to this:

class SearchForm(forms.Form):
    search = forms.CharField()

lessons: the field of the form needs to match with the keywords in the url. :)

answered Jan 8, 2023 at 3:56

chez93

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