How to return 404 and 500 errors in json formats instead of standard html pages in Django 4.x

lolz 0
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I found only very stale articles on the Internet that don't quite solve my problem.

I tried this way, but Django still returns html pages:

exceptions.py

def custom_exception_handler(exc, context):
    response = exception_handler(exc, context)

    if response is not None:
        if response.status_code == 500:
            response.data = {"code": 500, "message": "Internal server error"}
        if response.status_code == 404:
            response.data = {"code": 404, "message": "The requested resource was not found on this server"}
    return response

settings.py

REST_FRAMEWORK = {
    'EXCEPTION_HANDLER': 'backend.exceptions.custom_exception_handler',
}
django-rest-framework django
1 Answers

I think you'd just override the default templates..

I've override them using this in a view.py:

def handler404(request, exception):
    return render(request, 'custom_404.html', status=404)
def handler500(request):
    return render(request, 'custom_500.html', status=500)

That just seems to magically work, I really couldn't tell you how it knows to direct 500 to my handler.. it's not defined elsewhere in the project

and then for the template, custom_505.html, would look something like:

{{data}}
Nealium0
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