[Solved] Identify the error in this Django async view:async def my_view(request): result = slow_function() return JsonResponse({'data': result}) | Django

Django - Async Django

Identify the error in this Django async view:

async def my_view(request):
    result = slow_function()
    return JsonResponse({'data': result})

AMissing await before slow_function() call

BFunction should not be async

CJsonResponse cannot be returned from async views

DRequest parameter is missing

Step-by-Step Solution

Solution:

Step 1: Check async function calls
In async views, calling an async function requires await to pause until it finishes.
Step 2: Identify missing await
The code calls slow_function() without await, so it returns a coroutine object, not the result.
Final Answer:
Missing await before slow_function() call -> Option A
Quick Check:
Async calls need await = C [OK]

Quick Trick: Async calls must use await [OK]

Common Mistakes:

MISTAKES

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