Refactor existing code to use the new access module
Instead of ending up with many different ways to do
access control, we end up having only one centralized
place wher access control is done.
Patch by: Sverre Rabbelier
Reviewed by: Pawel Solyga, Augie Fackler, Todd Larsen
Reviewed at: http://codereviews.googleopensourceprograms.com/1601
Review id: 1601
from django.core.management.base import AppCommand
from django.utils.encoding import force_unicode
from django.utils.text import capfirst
MODULE_TEMPLATE = ''' {%% if perms.%(app)s.%(addperm)s or perms.%(app)s.%(changeperm)s %%}
<tr>
<th>{%% if perms.%(app)s.%(changeperm)s %%}<a href="%(app)s/%(mod)s/">{%% endif %%}%(name)s{%% if perms.%(app)s.%(changeperm)s %%}</a>{%% endif %%}</th>
<td class="x50">{%% if perms.%(app)s.%(addperm)s %%}<a href="%(app)s/%(mod)s/add/" class="addlink">{%% endif %%}Add{%% if perms.%(app)s.%(addperm)s %%}</a>{%% endif %%}</td>
<td class="x75">{%% if perms.%(app)s.%(changeperm)s %%}<a href="%(app)s/%(mod)s/" class="changelink">{%% endif %%}Change{%% if perms.%(app)s.%(changeperm)s %%}</a>{%% endif %%}</td>
</tr>
{%% endif %%}'''
class Command(AppCommand):
help = 'Prints the admin-index template snippet for the given app name(s).'
def handle_app(self, app, **options):
from django.db.models import get_models
output = []
app_models = get_models(app)
app_label = app_models[0]._meta.app_label
output.append('{%% if perms.%s %%}' % app_label)
output.append('<div class="module"><h2>%s</h2><table>' % app_label.title())
for model in app_models:
if model._meta.admin:
output.append(MODULE_TEMPLATE % {
'app': app_label,
'mod': model._meta.module_name,
'name': force_unicode(capfirst(model._meta.verbose_name_plural)),
'addperm': model._meta.get_add_permission(),
'changeperm': model._meta.get_change_permission(),
})
output.append('</table></div>')
output.append('{% endif %}')
return '\n'.join(output)