Adding Additional Data to a Serialize Response in Django - json

Question

Updated
I changed my simplified question into a real example.
I've created a working post response of data from the model using ModelSerialzer, which I call from a post method in a view class. I would like to add additional data to the response. This is the pertinent code from my CBV:
def post(self, request, format=None):
user_profile = UserProfiles.objects.get(user=request.user.id)
service_id = user_profile.service_id
rec_filter = Recommendations.objects.values_list('resource')
if service_id > 0:
service_name = Services.objects.get(pk=service_id)
programs = Programs.objects.filter(services=service_id)
resources_filtered = Resources.objects.filter(program__in=programs).exclude(id__in=rec_filter)
else:
service_name = 'All Services'
resources_filtered = Resources.objects.exclude(id__in=rec_filter)
serializer = ResourceSerializer(resources_filtered, many=True)
#serializer.data["service"] = service_name
return Response(serializer.data)
The commented out line was my attempt to add data base on a similar post here. I get a 500 response in my API call. What is the correct way to do it? The response data is JSON if that's necessary to mention.
This is the ModelSerializer:
class ResourceSerializer(serializers.ModelSerializer):
organization = OrganizationSerializer(read_only=True)
program = ProgramSerializer(read_only=True)
class Meta:
model = Resources
fields = [
'organization',
'program',
'link',
'contact',
'general_contact',
'eligibility',
'service_detail'
]
Test of the answer
Heres the updated code based on the answer with a correction to fix and error:
class ResourceSerializer(serializers.ModelSerializer):
organization = OrganizationSerializer(read_only=True)
program = ProgramSerializer(read_only=True)
service = serializers.SerializerMethodField()
def get_service(self, obj):
return "All Services"
class Meta:
model = Resources
fields = [
'organization',
'program',
'link',
'contact',
'general_contact',
'eligibility',
'service_detail',
'service'
]
The problem with this approach is that the value "All Services" is repeated in every row serialized. It's only needed once. I'd also like to keep the data transmitted minimized.

Answer 1

The problem with the original attempt is that serializer.data is immutable. It's necessary to make a copy and add to it.
serializer = ResourceSerializer(resources_filtered, many=True)
augmented_serializer_data = list(serializer.data)
augmented_serializer_data.append({'service': 'All Services'})
return Response(augmented_serializer_data)
This answer is based on one given by #andre-machado in this question.
This code here is an example to coincide with the other answer given.

Answer 2

You can do it in serializer itself. Define the new field required and add it in fields. Mark all the fields in serializer from resource model.
class ResourceSerializer(serializers.ModelSerializer):
service = serializers.SerializerMethodField()
def get_service(self):
return "All Services"
class Meta :
model = Resources
fields = ('service') #Mark all the fields required here from resource model

Answer 3

You can do it from the serilaizer. In this case i was adding the field isOpen to the response and this is how i did it .timeDifference is the name of the function that was to generate data for the extra field . I hope it helps
class ShopSearchSerializer(serializers.ModelSerializer):
isOpen = serializers.SerializerMethodField('timeDifference')
def timeDifference(self,*args):
requestTime = datetime.now()
return requestTime
class Meta:
model = Shop
fields =['name','city','street','house','opening_time','closing_time','isOpen']