27👍
This answer is probably arriving very late for you. But here it is.
You can use this in your Django code:
raise PermissionDenied("Custom message")
And then display the custom message using below snippet in the 403.html template:
{% if exception %}
<p>{{ exception }}</p>
{% else %}
<p>Static generic message</p>
{% endif %}
The message string passed to ‘PermissionDenied’ is available in template context as explained in Django documentation –
https://docs.djangoproject.com/en/1.10/ref/views/#http-forbidden-view
2👍
I came across the same issue and resolved it using the Django messages framework to pass a custom message to the template.
https://docs.djangoproject.com/en/1.8/ref/contrib/messages/
My specific example:
from django.contrib import messages
...
messages.error(request, 'The submission deadline has passed.')
raise PermissionDenied
The messages can then be output in the template as described in the documentation.
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1👍
You can try like this:
class SomeException(Exception):
message = 'An error occurred.'
def __init__(self, message):
self.message = message
def __str__(self):
return repr(self.message)
#usage
raise SomeException("Hello, you have an exception here")
Another way of sending a message to template is like:
if not request.user.is_staff: #or your condition
context['flash_message']= "permission error occurred"
retrun render_to_response('template.html', context)
# template
<!-- I am using bootstrap here -->
<div class="alert alert-{{ flash_message_type }} flash_message hide">
{{ flash_message | safe }}
</div>
<script>
...
if($.trim($(".flash_message").html()) != ''){
$(".flash_message").slideDown();
setTimeout(function(){
$(".flash_message").slideUp();
}, 5000);
};
</script>
1👍
If you are using class based views (CBV), or anything that extends the AccessMixin
, either set the permission_denied_message
attribute, or override the get_permission_denied_message
method.
Example:
from django.conf import settings
class MyView(ListView):
permission_denied_message = 'Hooo!'
def get_permission_denied_message(self):
perms = self.get_permission_required()
if settings.DEBUG:
for perm in perms:
if not self.request.user.has_perm(perm):
return 'Permission denied: ' + str(perm)
return ''
and then, in your template:
{% if exception and debug %}
<h3 class="font-bold">{{ exception }}</h3>
{% endif %}
0👍
Same issue here.
In django 1.9 we have this built in. In earliest django version we can use sys.exc_info()
, so the next step is just reuse whole default permission_denied
handler to add our exception.
# urls.py
...
handler403 = 'project.views.errors.permission_denied'
...
# views/errors.py
import sys
from django import http
from django.template import Context, TemplateDoesNotExist, loader
from django.views.decorators.csrf import requires_csrf_token
@requires_csrf_token
def permission_denied(request, template_name='403.html'):
_, value, _ = sys.exc_info()
try:
template = loader.get_template(template_name)
except TemplateDoesNotExist:
return http.HttpResponseForbidden('<h1>403 Forbidden</h1>', content_type='text/html')
return http.HttpResponseForbidden(
template.render(request=request, context={'exception': force_text(value)})
)
# templates/403.html
...
{{ exception }}
...