Add Permission to Django Admin

Question

Last Month i posted question on stackoverflow and on Django-Users group on G+ and on django website too. But i didn't find any answer that can solve my problem. What i want to do is to add new permission named as view in django admin panel, so user can only view data!. I also followed different patches from django website and tried django-databrowse but nothing works as expected. I then finally decide to edit views of auth/admin. Now what i am going to do is to add view permission like:

1. Added 'view' to default permission list

#./contrib/auth/management/init.py
def _get_all_permissions(opts):

    "Returns (codename, name) for all permissions in the given opts."
    perms = []
    for action in ('add', 'change', 'delete', 'view'):

        perms.append((_get_permission_codename(action, opts), u'Can %s %s' % (action, opts.verbose_name_raw)))

    return perms + list(opts.permissions)

2. Test the 'view' permission is added to all models

run manage.py syncdb

After this i can assign only view permission to user. Now this view permission must work too. So i am writing this code: in view.py of django-admin

for per in request.user.user_permissions_all():
    print per

This code prints permissions assigned to login user like auth | permission | can view department etc

Now i can get permission type and model name by splitting this sentence. I will get all the model name of application and will match that which data must b visible. This is again not what i really need but can work.

So my question is :

* Is this is what i should do or is there any other way too. I just want a solution that must works as expected. Need Your Assistance *

Answer 1

Adding 'view' permission to default permissions list

Your solution works, but you should really avoid editing source code if possible. There's a few ways to accomplish this within the framework:

1. Add the permission during post_syncdb():

In a file under your_app/management/

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)

Whenever you issue a 'syncdb' command, all content types can be checked to see if they have a 'view' permission, and if not, create one.

• SOURCE: The Nyaruka Blog

2. Add the permission to the Meta permissions option:

Under every model you would add something like this to its Meta options:

class Pizza(models.Model):
    cheesiness = models.IntegerField()

    class Meta:
        permissions = (
            ('view_pizza', 'Can view pizza'),
        )

This will accomplish the same as 1 except you have to manually add it to each class.

3. NEW in Django 1.7, Add the permission to the Meta default_permissions option:

In Django 1.7 they added the default_permissions Meta option. Under every model you would add 'view' to the default_permissions option:

class Pizza(models.Model):
    cheesiness = models.IntegerField()

    class Meta:
        default_permissions = ('add', 'change', 'delete', 'view')

---

Test the 'view' permission is added to all models

As for testing the whether a user has the permission, you can test on the has_perm() function. For example:

user.has_perm('appname.view_pizza') # returns True if user 'Can view pizza'