you are looking something like
https://code.djangoproject.com/ticket/7150
https://code.djangoproject.com/attachment/ticket/7150/admin_view-permission-1.0.patch
https://code.djangoproject.com/attachment/ticket/7150/newforms-admin-view-permission-r7737.patch
https://code.djangoproject.com/ticket/8936
https://code.djangoproject.com/ticket/22452
https://code.djangoproject.com/ticket/18814
https://code.djangoproject.com/ticket/17295
https://code.djangoproject.com/ticket/3984
https://code.djangoproject.com/ticket/2058
https://code.djangoproject.com/ticket/820
I don't know why Django doesn't support readonly view, various proposal was summit and always was reject. https://code.djangoproject.com/ticket/7150
https://code.djangoproject.com/attachment/ticket/7150/admin_view-permission-1.0.patch
https://code.djangoproject.com/attachment/ticket/7150/newforms-admin-view-permission-r7737.patch
https://code.djangoproject.com/ticket/8936
https://code.djangoproject.com/ticket/22452
https://code.djangoproject.com/ticket/18814
https://code.djangoproject.com/ticket/17295
https://code.djangoproject.com/ticket/3984
https://code.djangoproject.com/ticket/2058
https://code.djangoproject.com/ticket/820
2016-10-24 7:26 GMT-06:00 Derek <gamesbook@gmail.com>:
Would it be possible to add these extra admin methods into a parent class; and then have all your individual model admins inherit from it? Ditto for the models.py--
On Sunday, 8 February 2015 21:15:42 UTC+2, Hangloser Firestarter wrote:Solved.
In __init__.py
...
from django.db.models.signals import post_syncdb
from django.contrib.contenttypes.models import ContentType
from django.contrib.auth.models import Permission
def add_view_permissions(sender, **kwargs):
"""
This syncdb hooks takes care of adding a view permission too all our
content types.
"""
# for each of our content types
for content_type in ContentType.objects.all():
# build our permission slug
codename = "view_%s" % content_type.model
# if it doesn't exist..
if not Permission.objects.filter(content_type=content_type, codename=codename):
# add it
Permission.objects.create(content_type=content_type,
codename=codename,
name="Can view %s" % content_type.name)
print("Added view permission for %s" % content_type.name)
# check for all our view permissions after a syncdb
post_syncdb.connect(add_view_permissions)
...
In admin.py
...
class MyAdmin(admin.ModelAdmin):
def has_change_permission(self, request, obj=None):
ct = ContentType.objects.get_for_model(self.model)
salida = False
if request.user.is_superuser:
salida = True
else:
if request.user.has_perm('%s.view_%s' % (ct.app_label, ct.model)):
salida = True
else:
if request.user.has_perm('%s.change_%s' % (ct.app_label, ct.model)):
salida = True
else:
salida = False
return salida
def get_readonly_fields(self, request, obj=None):
ct = ContentType.objects.get_for_model(self.model)
if not request.user.is_superuser and request.user.has_perm('%s.view_%s' % (ct.app_label, ct.model)):
return [el.name for el in self.model._meta.fields]
return self.readonly_fields
...
in models.py
...
class City(models.Model):
nome_cidade = models.CharField(max_length=100)
estado_cidade = models.CharField(max_length=100)
pais_cidade = models.CharField(max_length=100)
def __str__(self):
return self.nome_cidade
class Meta:
permissions = (
('view_city', 'Can view city'),
)
...
Thank's for help!
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--
"La utopía sirve para caminar" Fernando Birri
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