发表新帖
发表新帖

How do I raise a Response Forbidden in django

前端 未结
关注
4 1122
情深已故
情深已故 2020-12-12 23:09

I\'d like to do the following:

raise HttpResponseForbidden()

But I get the error:

exceptions must be old-style classes or d
相关标签:
4条回答
  • 时光取名叫无心
    2020-12-12 23:33

    if you want to raise an exception you can use:

    from django.core.exceptions import PermissionDenied

def your_view(...):
    raise PermissionDenied()

    It is documented here :

    https://docs.djangoproject.com/en/stable/ref/views/#the-403-http-forbidden-view

    As opposed to returing HttpResponseForbidden, raising PermissionDenied causes the error to be rendered using the 403.html template, or you can use middleware to show a custom "Forbidden" view.

    0 讨论(0)
  • 小蘑菇
    2020-12-12 23:39

    You can optionally supply a custom template named "403.html" to control the rendering of 403 HTTP errors.

    As correctly pointed out by @dave-halter, The 403 template can only be used if you raise PermissionDenied

    Below is a sample view used to test custom templates "403.html", "404.html" and "500.html"; please make sure to set DEBUG=False in project's settings or the framework will show a traceback instead for 404 and 500.

    from django.http import HttpResponse
from django.http import Http404
from django.core.exceptions import PermissionDenied


def index(request):

    html = """
<!DOCTYPE html>
<html lang="en">
  <body>
    <ul>
        <li><a href="/">home</a></li>
        <li><a href="?action=raise403">Raise Error 403</a></li>
        <li><a href="?action=raise404">Raise Error 404</a></li>
        <li><a href="?action=raise500">Raise Error 500</a></li>
    </ul>
  </body>
</html>
"""

    action = request.GET.get('action', '')
    if action == 'raise403':
        raise PermissionDenied
    elif action == 'raise404':
        raise Http404
    elif action == 'raise500':
        raise Exception('Server error')

    return HttpResponse(html)
    0 讨论(0)
  • 北荒
    2020-12-12 23:43

    Return it from the view as you would any other response.

    from django.http import HttpResponseForbidden

return HttpResponseForbidden()
    0 讨论(0)
  • 滥情空心
    2020-12-12 23:51

    Try this Way , sending message with Error 

    from django.core.exceptions import PermissionDenied
raise PermissionDenied("You do not have permission to Enter Clients in Other Company, Be Careful")
    0 讨论(0)
 看不清?
提交回复
热议问题