Add permission_is_exempt()

This commit is contained in:
Jeremy Stretch 2020-06-01 12:31:18 -04:00
parent 3c334a0238
commit 9679557747

View File

@ -1,5 +1,5 @@
from django.conf import settings
from django.contrib.contenttypes.models import ContentType
from django.db.models import Q
def get_permission_for_model(model, action):
@ -34,3 +34,25 @@ def resolve_permission(name):
raise ValueError(f"Unknown app/model for {name}")
return content_type, action
def permission_is_exempt(name):
"""
Determine whether a specified permission is exempt from evaluation.
:param name: Permission name in the format <app>.<action>_<model>
"""
app_label, codename = name.split('.')
action, model_name = codename.split('_')
if action == 'view':
if (
# All models are exempt from view permission enforcement
'*' in settings.EXEMPT_VIEW_PERMISSIONS
) or (
# This specific model is exempt from view permission enforcement
'{}.{}'.format(app_label, model_name) in settings.EXEMPT_VIEW_PERMISSIONS
):
return True
return False