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Updated permissions app to only return active permissions on the views and querysets

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This commit is contained in:
Roberto Rosario
2011-10-29 02:23:33 -04:00
parent 0c512ccaec
commit e048711d82
3 changed files with 20 additions and 11 deletions
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8
apps/permissions/views.py
View File

@@ -22,7 +22,7 @@ from permissions.forms import RoleForm, RoleForm_view
from permissions import PERMISSION_ROLE_VIEW, PERMISSION_ROLE_EDIT, \
PERMISSION_ROLE_CREATE, PERMISSION_ROLE_DELETE, PERMISSION_PERMISSION_GRANT, \
PERMISSION_PERMISSION_REVOKE
from permissions.api import check_permissions, namespace_titles
from permissions.api import check_permissions, namespace_titles, get_permission_label, get_permission_namespace_label
from permissions.widgets import role_permission_link
@@ -52,10 +52,10 @@ def role_permissions(request, role_id):
'name': u'generic_list_subtemplate.html',
'context': {
'title': _(u'permissions'),
'object_list': Permission.objects.all(),
'object_list': Permission.objects.active_only(),
'extra_columns': [
{'name': _(u'namespace'), 'attribute': encapsulate(lambda x: namespace_titles[x.namespace] if x.namespace in namespace_titles else x.namespace)},
{'name': _(u'name'), 'attribute': u'label'},
{'name': _(u'namespace'), 'attribute': encapsulate(lambda x: get_permission_namespace_label(x))},
{'name': _(u'name'), 'attribute': encapsulate(lambda x: get_permission_label(x))},
{
'name':_(u'has permission'),
'attribute': encapsulate(lambda x: two_state_template(x.has_permission(role))),