Returning JSON with JsonResponse in Django - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the correct class for returning JSON responses in Django.

Django

from django.http import [1]

Drag options to blanks, or click blank then click option'

AHttpResponse

BJsonResponse

CHttpRequest

DJsonRequest

Attempts:

3 left

💡 Hint

Common Mistakes

Using HttpResponse instead of JsonResponse.

Confusing JsonResponse with HttpRequest.

The JsonResponse class is used to return JSON data in Django views.

2fill in blank

medium

Complete the code to return a JSON response with a dictionary containing a message.

Django

def my_view(request):
    data = {'message': 'Hello, world!'}
    return [1](data)

Drag options to blanks, or click blank then click option'

Aredirect

BHttpResponse

CJsonResponse

Drender

Attempts:

3 left

💡 Hint

Common Mistakes

Using HttpResponse without setting content type.

Using render which is for HTML templates.

Use JsonResponse to return JSON data from a Django view.

3fill in blank

hard

Fix the error in the code to correctly return JSON with safe=False for non-dict data.

Django

def my_view(request):
    data = ['apple', 'banana', 'cherry']
    return JsonResponse(data, [1]=False)

Drag options to blanks, or click blank then click option'

Asafe

Bjson_safe

Callow_non_dict

Dsafe_mode

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect parameter names like json_safe or safe_mode.

Not setting safe=False when returning a list.

The safe parameter must be set to False to allow non-dictionary objects in JsonResponse.

4fill in blank

hard

Fill both blanks to create a JSON response with a custom status code and headers.

Django

def my_view(request):
    data = {'status': 'ok'}
    return JsonResponse(data, status=[1], headers=[2])

Drag options to blanks, or click blank then click option'

A201

B404

C{'X-Custom-Header': 'Value'}

D{'Content-Type': 'application/json'}

Attempts:

3 left

💡 Hint

Common Mistakes

Using status codes like 404 for success.

Passing headers as a string instead of a dictionary.

Use status=201 for created response and pass headers as a dictionary.

5fill in blank

hard

Fill all three blanks to return a JSON response with a nested dictionary, custom status, and safe=False.

Django

def my_view(request):
    data = [1]
    return JsonResponse(data, status=[2], safe=[3])

Drag options to blanks, or click blank then click option'

A{'items': ['a', 'b', 'c']}

B202

CFalse

D['a', 'b', 'c']

Attempts:

3 left

💡 Hint

Common Mistakes

Passing a list directly without safe=False.

Using incorrect status codes.

Not matching data type with safe parameter.

The data is a dictionary with a list inside, status 202 means accepted, and safe=False allows non-dict if needed.

Practice

1. What is the main purpose of Django's JsonResponse class?

easy

A. To send JSON data as an HTTP response easily and safely

B. To render HTML templates with JSON data

C. To parse JSON data from incoming requests

D. To convert JSON strings into Python dictionaries

Solution

1. Step 1: Understand JsonResponse role

   JsonResponse is designed to send JSON data back to the client as an HTTP response.

2. Step 2: Differentiate from other JSON tasks

   It does not parse incoming JSON or render HTML; it only sends JSON safely with correct headers.

3. Final Answer:

   To send JSON data as an HTTP response easily and safely -> Option A

4. Quick Check:

   JsonResponse sends JSON response [OK]

Hint: JsonResponse sends JSON data back to clients [OK]

Common Mistakes:

• Confusing JsonResponse with JSON parsing
• Thinking it renders HTML templates
• Assuming it converts JSON strings to dicts

2. Which of the following is the correct way to return a JSON response with a dictionary {"name": "Alice"} using JsonResponse?

easy

A. return JsonResponse([{"name": "Alice"}], safe=False)

B. return JsonResponse({"name": "Alice"})

C. return JsonResponse('[{"name": "Alice"}]')

D. return JsonResponse(json.dumps({"name": "Alice"}))

Solution

1. Step 1: Use JsonResponse with a dictionary

   JsonResponse accepts a dictionary by default and sets safe=True, so no extra argument is needed.

2. Step 2: Differentiate correct usage

   Passing a list of dictionaries produces a JSON array instead of a single object; passing JSON strings causes TypeError.

3. Final Answer:

   return JsonResponse({"name": "Alice"}) -> Option B

4. Quick Check:

   Dict input with default safe=True [OK]

Hint: Pass dict directly; no safe=False needed for dict [OK]

Common Mistakes:

• Passing JSON string instead of dict
• Passing a list produces JSON array instead of object
• Using json.dumps inside JsonResponse

3. What will be the output of this Django view code?

from django.http import JsonResponse

def my_view(request):
    data = [1, 2, 3]
    return JsonResponse(data, safe=False)

medium

A. HTTP response with JSON array [1, 2, 3]

B. TypeError because safe=False is invalid

C. HTTP response with JSON string '[1, 2, 3]'

D. HTTP response with JSON object {"data": [1, 2, 3]}

Solution

1. Step 1: Recognize safe=False allows list data

   JsonResponse by default requires dict; safe=False allows lists like [1, 2, 3].

2. Step 2: Understand output JSON content

   The response will contain the JSON array [1, 2, 3] as the body with correct headers.

3. Final Answer:

   HTTP response with JSON array [1, 2, 3] -> Option A

4. Quick Check:

   safe=False lets list output [OK]

Hint: Use safe=False to return lists as JSON [OK]

Common Mistakes:

• Omitting safe=False causes TypeError
• Expecting JSON object instead of array
• Confusing JSON string with JSON array

4. Identify the error in this Django view code:

from django.http import JsonResponse

def my_view(request):
    data = ["apple", "banana"]
    return JsonResponse(data)

medium

A. No error; code works fine

B. JsonResponse cannot return lists at all

C. JsonResponse requires json.dumps before returning

D. Missing safe=False causes TypeError for list data

Solution

1. Step 1: Check data type passed to JsonResponse

   Data is a list, but JsonResponse defaults to safe=True, which only allows dicts.

2. Step 2: Understand error cause

   Without safe=False, passing a list raises TypeError because it's unsafe to send raw lists.

3. Final Answer:

   Missing safe=False causes TypeError for list data -> Option D

4. Quick Check:

   List needs safe=False to avoid error [OK]

Hint: Add safe=False when returning lists [OK]

Common Mistakes:

• Forgetting safe=False with lists
• Trying to json.dumps before JsonResponse
• Assuming JsonResponse accepts lists by default

5. You want to return a JSON response with a list of user names ["Alice", "Bob"] from a Django view. Which is the correct way to do this?

hard

A. return JsonResponse({"users": ["Alice", "Bob"]})

B. return JsonResponse(["Alice", "Bob"], safe=True)

C. return JsonResponse(["Alice", "Bob"], safe=False)

D. return JsonResponse(json.dumps(["Alice", "Bob"]))

Solution

1. Step 1: Decide JSON structure to return

   Returning a list directly requires safe=False because JsonResponse defaults to safe=True (dict only).

2. Step 2: Avoid common mistakes

   Passing safe=True with a list causes error; json.dumps returns string, which JsonResponse does not accept.

3. Final Answer:

   return JsonResponse(["Alice", "Bob"], safe=False) -> Option C

4. Quick Check:

   List with safe=False returns JSON array [OK]

Hint: Use safe=False to return JSON lists directly [OK]

Common Mistakes:

• Setting safe=True with list causes error
• Passing JSON string instead of Python list
• Wrapping list inside dict unnecessarily