[Solved] How would you modify this Django view to handle invalid JSON in the request body gracefully?from django.http import JsonResponse import json def view(request): data = json.loads(request.... | Django

Django - REST Framework Basics

How would you modify this Django view to handle invalid JSON in the request body gracefully?

from django.http import JsonResponse
import json

def view(request):
    data = json.loads(request.body)
    return JsonResponse({'received': data})

AWrap json.loads in try-except block catching json.JSONDecodeError and return error JsonResponse

BUse request.POST instead of json.loads to avoid errors

CAdd content_type='application/json' to JsonResponse call

DUse json.dumps instead of json.loads

Step-by-Step Solution

Solution:

Step 1: Identify error source
json.loads can raise JSONDecodeError if request.body is invalid JSON.
Step 2: Add try-except to catch error and respond gracefully
Wrap parsing in try-except and return JsonResponse with error message on exception.
Final Answer:
Wrap json.loads in try-except catching JSONDecodeError -> Option A
Quick Check:
Use try-except to handle JSON parsing errors [OK]

Quick Trick: Catch JSONDecodeError to handle invalid JSON gracefully [OK]

Common Mistakes:

MISTAKES

Master "REST Framework Basics" in Django

9 interactive learning modes - each teaches the same concept differently

More Django Quizzes

How would you modify this Django view to handle invalid JSON in the request body gracefully?