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How to raise a JSON:API exception with a custom status code?#1128

Answeredbysliverc
Nekidev asked this question inQ&A
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What the title says. For example, I can raise aValidationError withforbidden as the code, but the status code will be 400 (not 403).

fromrest_framework_json_apiimportserializersdefmy_error_view(request):"""    This view raises an exception which will be formatted as a JSON:API error object.    """raiseserializers.ValidationError(detail="You need the `access_this_view` permission to access this resource.",code="forbidden"    )

Any request to that endpoint in this example will return a 400 error response with aforbidden code. Is there any way to specify a status code?

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Django REST framework has different type ofexceptions for different status codes.ValidationError will always return a status code 400. In case you want to have 403 status code you need to raise aPermissionDenied exception.

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Django REST framework has different type ofexceptions for different status codes.ValidationError will always return a status code 400. In case you want to have 403 status code you need to raise aPermissionDenied exception.

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