Request parsing and response rendering in Django - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ component_behavior

intermediate

What is the output of this Django view when receiving a POST request with JSON data?

Consider this Django view that parses JSON data from a POST request and returns a JSON response:

from django.http import JsonResponse
import json

def my_view(request):
    if request.method == 'POST':
        data = json.loads(request.body.decode('utf-8'))
        name = data.get('name', 'Guest')
        return JsonResponse({'greeting': f'Hello, {name}!'})
    return JsonResponse({'error': 'Only POST allowed'}, status=405)

What will be the JSON response body if the POST request body is {"name": "Alice"}?

Django

from django.http import JsonResponse
import json

def my_view(request):
    if request.method == 'POST':
        data = json.loads(request.body.decode('utf-8'))
        name = data.get('name', 'Guest')
        return JsonResponse({'greeting': f'Hello, {name}!'})
    return JsonResponse({'error': 'Only POST allowed'}, status=405)

A {"greeting": "Hello, Alice!"}

B {"greeting": "Hello, Guest!"}

C {"error": "Only POST allowed"}

D SyntaxError

💡 Hint

Look at how the JSON body is parsed and how the 'name' key is used.

The view loads the JSON from request.body, extracts the 'name' key, and returns a greeting with that name. Since the POST body contains {"name": "Alice"}, the greeting is "Hello, Alice!".

📝 Syntax

intermediate

Which option correctly parses JSON data in a Django view?

Which of the following code snippets correctly parses JSON data from a Django request object?

A data = request.json()

B data = request.POST.json()

C data = json.loads(request.body.decode('utf-8'))

D data = json.parse(request.body)

💡 Hint

Remember that request.body is a byte string and you need to decode it properly.

In Django, request.body contains the raw bytes of the request body. Using json.loads(request.body.decode('utf-8')) correctly parses the JSON. The other options are invalid because Django's request.POST is for form data, and request has no json() method. Also, json.parse does not exist in Python.

🔧 Debug

advanced

Why does this Django view raise a TypeError when returning JsonResponse?

Examine this Django view:

from django.http import JsonResponse

def my_view(request):
    data = {'count': 5, 'items': [1, 2, 3]}
    return JsonResponse(data, safe=False)

Why does this code raise a TypeError?

Django

from django.http import JsonResponse

def my_view(request):
    data = {'count': 5, 'items': [1, 2, 3]}
    return JsonResponse(data, safe=False)

A Because safe=False is required only when returning a list, not a dict

B Because safe=False is used with a dict, which is not allowed

C Because JsonResponse cannot serialize lists inside dicts

D Because the data dict is missing a status code

💡 Hint

Check the meaning of the safe parameter in JsonResponse.

The safe parameter in JsonResponse must be set to False only when the top-level object is not a dict (e.g., a list). Since data is a dict here, safe=False is unnecessary but does not cause an error. Actually, this code runs fine. The question is tricky because the code does not raise a TypeError. The correct answer is that safe=False is only needed for non-dict objects.

❓ state_output

advanced

What is the HTTP status code of this Django JsonResponse?

Given this Django view:

from django.http import JsonResponse

def error_view(request):
    return JsonResponse({'error': 'Not found'}, status=404)

What HTTP status code will the response have?

Django

from django.http import JsonResponse

def error_view(request):
    return JsonResponse({'error': 'Not found'}, status=404)

A 500

B 404

C 200

D 400

💡 Hint

Look at the status parameter in JsonResponse.

The status parameter sets the HTTP status code of the response. Here it is explicitly set to 404, so the response will have status code 404.

🧠 Conceptual

expert

Which Django feature automatically parses JSON request bodies in views?

In Django, which feature or tool automatically parses JSON request bodies and provides the data as a Python dictionary without manually calling json.loads(request.body)?

A Using Django's HttpRequest.POST attribute

B Using Django's middleware for JSON parsing

C Using Django's built-in JsonResponse

D Using Django REST Framework's Request object

💡 Hint

Think about popular Django extensions for APIs.

Django itself does not automatically parse JSON request bodies. However, Django REST Framework (DRF) provides a Request object that parses JSON and other content types automatically, making the data available as request.data. The other options either do not parse JSON automatically or are unrelated.

Practice

1. What is the main purpose of JsonResponse in Django?

easy

A. To send JSON data back to the client as an HTTP response

B. To parse JSON data from the client's request body

C. To convert Python objects into HTML templates

D. To handle file uploads in a Django view

Solution

1. Step 1: Understand JsonResponse role

   JsonResponse is a Django class that formats Python data as JSON and sends it as an HTTP response.

2. Step 2: Differentiate from request parsing

   Parsing JSON from requests is done with json.loads() or similar, not JsonResponse.

3. Final Answer:

   To send JSON data back to the client as an HTTP response -> Option A

4. Quick Check:

   JsonResponse sends JSON responses [OK]

Hint: JsonResponse sends data out, not reads it in [OK]

Common Mistakes:

• Confusing JsonResponse with JSON parsing
• Thinking JsonResponse parses request data
• Mixing up response rendering with template rendering

2. Which of the following is the correct way to parse JSON data from a Django request object named request?

easy

A. data = request.json()

B. data = json.dumps(request.body)

C. data = JsonResponse(request.body)

D. data = json.loads(request.body)

Solution

1. Step 1: Identify JSON parsing method

   To convert JSON string to Python object, use json.loads().

2. Step 2: Apply to request body

   request.body contains raw bytes, so decode if needed, then parse with json.loads().

3. Final Answer:

   data = json.loads(request.body) -> Option D

4. Quick Check:

   json.loads parses JSON string [OK]

Hint: Use json.loads to read JSON from request body [OK]

Common Mistakes:

• Using json.dumps instead of json.loads
• Calling non-existent request.json() method
• Using JsonResponse to parse input

3. Given this Django view code, what will be the HTTP response content?

from django.http import JsonResponse
import json

def my_view(request):
    data = json.loads(request.body)
    result = {"message": f"Hello, {data['name']}!"}
    return JsonResponse(result)

And the client sends JSON body: {"name": "Alice"}

medium

A. Hello, Alice!

B. {"message": "Hello, {data['name']}!"}

C. {"message": "Hello, Alice!"}

D. SyntaxError

Solution

1. Step 1: Parse JSON from request body

   The code uses json.loads(request.body) to get a Python dict with key 'name' and value 'Alice'.

2. Step 2: Format message and return JsonResponse

   The message string becomes "Hello, Alice!" and is wrapped in a dict, then sent as JSON response.

3. Final Answer:

   {"message": "Hello, Alice!"} -> Option C

4. Quick Check:

   JsonResponse sends formatted JSON string [OK]

Hint: JsonResponse returns JSON string with keys and values [OK]

Common Mistakes:

• Thinking JsonResponse returns plain text
• Confusing string interpolation syntax
• Expecting raw Python dict as response

4. What is wrong with this Django view code snippet for parsing JSON?

def my_view(request):
    data = json.loads(request.POST)
    return JsonResponse({"status": "ok"})

medium

A. request.POST is empty for GET requests only

B. request.POST is not a JSON string, so json.loads will fail

C. The function must return a string, not JsonResponse

D. JsonResponse cannot be used without importing

Solution

1. Step 1: Understand request.POST content

   request.POST is a QueryDict, not a JSON string, so passing it to json.loads() causes an error.

2. Step 2: Correct JSON parsing method

   To parse JSON, use json.loads(request.body) instead, since request.body contains raw JSON bytes.

3. Final Answer:

   request.POST is not a JSON string, so json.loads will fail -> Option B

4. Quick Check:

   json.loads needs JSON string, not QueryDict [OK]

Hint: Use request.body for JSON, not request.POST [OK]

Common Mistakes:

• Passing request.POST to json.loads
• Assuming JsonResponse needs string return
• Ignoring HTTP method differences

5. You want to create a Django view that accepts JSON with a list of numbers under key values, sums them, and returns the sum as JSON. Which code correctly implements this?

hard

A. def sum_view(request):
data = json.loads(request.body)
total = sum(data['values'])
return JsonResponse({'sum': total})

B. def sum_view(request):
total = sum(request.POST.getlist('values'))
return JsonResponse({'sum': total})

C. def sum_view(request):
data = json.dumps(request.body)
total = sum(data['values'])
return JsonResponse({'sum': total})

D. def sum_view(request):
total = sum(request.GET['values'])
return JsonResponse({'sum': total})

Solution

1. Step 1: Parse JSON body correctly

   Use json.loads(request.body) to get Python dict from JSON input.

2. Step 2: Sum the list and return JSON response

   Extract the list under 'values', sum it, and return with JsonResponse.

3. Final Answer:

   def sum_view(request): data = json.loads(request.body) total = sum(data['values']) return JsonResponse({'sum': total}) -> Option A

4. Quick Check:

   Parse JSON body, sum list, return JsonResponse [OK]

Hint: Parse JSON body, then sum list, then JsonResponse [OK]

Common Mistakes:

• Using json.dumps instead of json.loads
• Trying to sum QueryDict values directly
• Using request.GET or request.POST for JSON body