有处理程序和异常到响应的转换器，它允许你执行raise Http404。正如你所看到的，它们也将PermissionDenied异常转换为带有403状态代码的响应。所以你可以引发 it 而不是Http403。但是对于401，你必须返回类似return HttpResponse('Unauthorized', status=401)的东西。

下面是Django处理异常的代码：

def response_for_exception(request, exc):
    if isinstance(exc, Http404):
        if settings.DEBUG:
            response = debug.technical_404_response(request, exc)
        else:
            response = get_exception_response(request, get_resolver(get_urlconf()), 404, exc)

    elif isinstance(exc, PermissionDenied):
        logger.warning(
            'Forbidden (Permission denied): %s', request.path,
            extra={'status_code': 403, 'request': request},
        )
        response = get_exception_response(request, get_resolver(get_urlconf()), 403, exc)

    elif isinstance(exc, MultiPartParserError):
        logger.warning(
            'Bad request (Unable to parse request body): %s', request.path,
            extra={'status_code': 400, 'request': request},
        )
        response = get_exception_response(request, get_resolver(get_urlconf()), 400, exc)

    elif isinstance(exc, SuspiciousOperation):
        if isinstance(exc, (RequestDataTooBig, TooManyFieldsSent)):
            # POST data can't be accessed again, otherwise the original
            # exception would be raised.
            request._mark_post_parse_error()

        # The request logger receives events for any problematic request
        # The security logger receives events for all SuspiciousOperations
        security_logger = logging.getLogger('django.security.%s' % exc.__class__.__name__)
        security_logger.error(
            force_text(exc),
            extra={'status_code': 400, 'request': request},
        )
        if settings.DEBUG:
            response = debug.technical_500_response(request, *sys.exc_info(), status_code=400)
        else:
            response = get_exception_response(request, get_resolver(get_urlconf()), 400, exc)

    elif isinstance(exc, SystemExit):
        # Allow sys.exit() to actually exit. See tickets #1023 and #4701
        raise

    else:
        signals.got_request_exception.send(sender=None, request=request)
        response = handle_uncaught_exception(request, get_resolver(get_urlconf()), sys.exc_info())

    # Force a TemplateResponse to be rendered.
    if not getattr(response, 'is_rendered', True) and callable(getattr(response, 'render', None)):
        response = response.render()

    return response

正如你在这里看到的，Django会处理Http404和PermissionDenied异常，而Http 401响应似乎没有异常。
PermissionDenied（Http 403）异常为以下异常（类似于Http404异常）：

class PermissionDenied(Exception):
    """The user did not have permission to do that"""
    pass

所以它应该可以和raise PermissionDenied一起工作。

**Django文档：**错误视图

您使用的是django rest framework吗？您可以像这样从APIException继承：

from rest_framework.exceptions import APIException

class UnauthorizedException(APIException):
    status_code = 401
    default_detail = "Not logged in"

比如在你的视野里或者任何你想去的地方

if request.user.is_anonymous:
        raise UnauthorizedException

打电话的人会得到401。