diff options
| author | Tim Graham <timograham@gmail.com> | 2013-07-11 11:06:34 -0400 |
|---|---|---|
| committer | Tim Graham <timograham@gmail.com> | 2013-07-11 11:07:59 -0400 |
| commit | c278d81da40430f401cb7ce24f6bc603c40f3de7 (patch) | |
| tree | 883c58ee83f827a3f871ebdb4db777d6545aeada /docs | |
| parent | 2f9378223d0c92447e2529d5eeaad7d301b78b84 (diff) | |
[1.5.x] Fixed #20730 -- Fixed "Programmatically creating permissions" error.
Thanks glarrain for the report.
Backport of 684a606a4e from master
Diffstat (limited to 'docs')
| -rw-r--r-- | docs/topics/auth/default.txt | 3 |
1 files changed, 2 insertions, 1 deletions
diff --git a/docs/topics/auth/default.txt b/docs/topics/auth/default.txt index 4072c51771..3234b53e55 100644 --- a/docs/topics/auth/default.txt +++ b/docs/topics/auth/default.txt @@ -234,10 +234,11 @@ a model's ``Meta`` class, you can also create permissions directly. For example, you can create the ``can_publish`` permission for a ``BlogPost`` model in ``myapp``:: + from myapp.models import BlogPost from django.contrib.auth.models import Group, Permission from django.contrib.contenttypes.models import ContentType - content_type = ContentType.objects.get(app_label='myapp', model='BlogPost') + content_type = ContentType.objects.get_for_model(BlogPost) permission = Permission.objects.create(codename='can_publish', name='Can Publish Posts', content_type=content_type) |
