Django role based views?
Asked Answered
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9

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I'm looking for some input on how others would architect this. I'm going to provide class (django group) based views.

For example, a user's group will determine what views/templates he or she will have access to. I'm thinking of perhaps storing paths to view functions in a table to determine what a user's link bar will consist of. Filter specifications can also be stored to determine what rows will fill these templates.

A good example is a hospital nursing units. Nurses at one unit need not see the entire hospital's patients. They only need to see their patients. Doctors on the same unit need only to see those patients as well, but they should have access to much greater functionality.

Has this been done via some third party application? And how would you approach this problem?

Thanks, Pete

Suasion answered 10/10, 2009 at 0:29 Comment(1)
Guys, the django permission system does not fit my needs. That's why I'm asking for architecture help :PSuasion
C
49

Django already has a groups and permissions system, which may be sufficient for your purpose.

http://docs.djangoproject.com/en/dev/topics/auth/

Generally in your code you check if a user has a permission. A user has his own permissions and those of the groups he belongs to. You can administer this pretty easily from the admin console.

There are two parts you need to look at.

  1. Check that a user requesting a page has permission to do so.
  2. Only display links to the user if he has the permission.

For 1. you can check permissions in a decorator as such:

from django.contrib.auth.decorators import permission_required

@permission_required('polls.can_vote')
def some_view(request):

For 2. the currently logged-in user's permissions are stored in the template variable {{ perms }}. This code checks the same permission as above.

{% if perms.polls.can_vote %}
    <a href="/vote">vote</a>
{% endif %}

To generate a list of links you can iterate over user.get_all_permissions() and fetch the links (or function that generates the link) from a dict:

def more_elaborate_list_of_links_for_a_perm(user):
    return ["/link1", ...]

_LINKS = {
    'polls.can_vote' : lambda u: ["/user/specific/link/" + u.id],
    'polls.can_close': lambda u: ['/static/link/1', 'static/link/2'],
    'polls.can_open' : more_elaborate_list_of_links_for_a_perm
}

def gen_links(user):
    # get_all_permissions also gets permissions for users groups
    perms = user.get_all_permissions()
    return sum((_LINKS[p](user) for p in perms if p in _LINKS), [])

There are probably many other approaches.

Cychosz answered 10/10, 2009 at 1:0 Comment(2)
How would you generate a list of links to provide a user using django's built-in permissions system?Suasion
For the record, I did not select this answer as the best. I don't think it is.Suasion
I
5

We had a similar problem. Django's groups aren't REALLY suited to this, but you can shoehorn them in.

The way we did it was as follows:

Every access-controlled object has a ManyToMany relation to the groups table. Each group was used to define a specific type of permission ("can view patient basics", "can edit patient contact info", and so on). Users are added to the groups that they should have permissions for (in your example of seeing only patients in this hospital, you could have a "valley-view-hospital" group).

Then when you go to display a list of records to a user, you filter based on the conjunction of the two groups. A user has to have all the associated group permissions to view a given object.

If your system requires it, you can keep a separate ManyToMany of negative permissions, or separate read/write permissions. You could also define a set of meta-groups (doctor, nurse) that result in your lookup filter retrieving the actual subset of permissions.

As far as your link-bar problem goes, you can generate those programatically using the same system - filter based on the classes of objects the user can see or edit, and then use a get_absolute_url() type function (maybe call it get_index_url()) to return the links for the index of each class of object.

Because all this is fairly complex, you'll probably end up wanting to do some level of caching for these things, but get it working before you bother optimizing. It is possible, and it's less ugly in code than it is in words.

Internationalism answered 29/10, 2009 at 6:1 Comment(2)
I've ended up going this route with our app, but its not ideal. I found performance has dropped a little. I'm looking elsewhere for a better ACL solutionTirzah
This is not scalable at all. Any ideas on ACL ?Baulk
S
4

There is a new very interesting project about role based permissions in Django: http://bitbucket.org/nabucosound/django-rbac

Shyamal answered 2/11, 2009 at 16:39 Comment(0)
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3

I had a similar problem not too long ago. Our solution did the trick, though it might be too simple for your situation. Like everyone is suggesting, we used the django permission system to control what user interactions with models. However, we didn't just try to group users, we also grouped objects through a GenericForeignKey.

We built a model that linked to itself to allow for hierarchies to be developed.

class Group( models.Model ):
    name = models.CharField( ... )
    parent = models.ForeignKey( 'self', blank=True, null=True)
    content_type = models.ForeignKey( ContentType )
    object_id = models.PositiveIntegerField()
    content_object = generic.GenericForeignKey( 'content_type', 'object_id' )
    ...

To make it work, we also created a model to serve as the django User model's user profile. All it contained was a ManyToManyField linked to the Group model above. This allowed us to give users access to zero or more Groups as required. (documentation)

class UserProfile( models.Model ):
    user = models.ForeignKey( User, unique=True )
    groups = models.ManyToManyField( Group )
    ...

This gave us the best of both worlds and kept us from trying to shoehorn everything into django's permission system. I'm using this basic setup to control user's access to sports content (some users can access whole leagues, some only one or two conferences, some only have access to individual teams), and it works well in that situation. It could probably be a generalized enough to fit your needs.

Ironwood answered 30/10, 2009 at 4:23 Comment(0)
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1

If you don't need real per-object ACLs, then you can just use the Django permission system. To get a list of all available permissions:

from django.contrib.auth.models import Permission
perms = Permission.objects.all()

There is an API for other authentication and authorization sources, so you do not need to stick with this permissions table.

You may hack this Django system to fit your needs in terms of this authorization model (RBAC) or you may come up with an ACL-like solution.

Xylon answered 10/10, 2009 at 2:4 Comment(1)
any hint for a per-object acl ?Mei
A
1

On a site for an expert on Pinot Noir wine we created per-object access based on a number of different criteria. If the inbound link had a referer field that matched the domain name of a featured winery, then the user got a 'winery token' which expanded to all articles, tasting notes, etc. related to that winery. We use 'named tokens' for give aways at tasting events and they gave access to specific parts of the site. We even use this to grant certain types of permissions to search engine spiders and then make sure that links that come from those search engines have the same permissions as the spider did (ie. no cloaking games).

The short version is that you can create a class (we called them TokenBuckets which hold Tokens) and each object (on a detail page, or a list page, or whatever) can ask the user's TokenBucket if a certain level of access is allowed.

Basically it's a weird kind of ACL system. It wasn't that hard to create the mechanics. All of the magic is in determining under what circumstances which tokens go into the bucket.

Aerify answered 30/10, 2009 at 4:37 Comment(0)
M
1

This question has been asked in Oct 2009 and the problem still exists in July 2012.

I have searched for a good Role-Based app, and found django-permission as the best result.

Three important features that I needed were Roles, view Decorators and Templatetag; apparently django-permissions has all of them. Read it's docs for it's usage.

The only drawback is that it's under development.

Masseter answered 4/7, 2012 at 8:22 Comment(0)
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1

You can use django user roles

https://github.com/dabapps/django-user-roles

Grubb answered 7/9, 2012 at 13:59 Comment(0)
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0

We used a role base system for a similar problem. Basically users have permissions to assume different roles.

View functions got decorated:

def needs_capability(capability,redirect_to="/cms/"):
   def view_func_wrapper(view_func):
       def wrapped_view_func(request,*args,**kwargs):
           if not request.role._can(capability):
              return HttpResponseRedirect(redirect_to)
           return view_func(request,*args,**kwargs)
       return wrapped_view_func
   return view_func_wrapper

The rest of the magic is inside the request.role attribute which got set inside a context processor. Authenticated users got a Role, for the unwashed masses a DummyRole.

Access to information was restricted further inside the templates:

 {% if not request.role.can.view_all_products %}
          Lots of products, yeah!
 {% endif %}

Not the cleanest solution in my opinion, but worked as expected.

Tierell answered 2/11, 2009 at 12:43 Comment(0)

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