Django role based views?

Question

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

Answer 1

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.