Any request to that endpoint in this example will return a 400 error response with a forbidden code. Is there any way to specify a status code?
Django REST framework has different type of exceptions 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 a PermissionDenied exception.
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What the title says. For example, I can raise a from rest_framework_json_api import serializers
def my_error_view(request):
"""
This view raises an exception which will be formatted as a JSON:API error object.
"""
raise serializers.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 a |
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Django REST framework has different type of exceptions for different status codes. |
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Django REST framework has different type of exceptions for different status codes.
ValidationErrorwill always return a status code 400. In case you want to have 403 status code you need to raise a PermissionDenied exception.