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    <p>Django already has a groups and permissions system, which may be sufficient for your purpose.</p> <p><a href="http://docs.djangoproject.com/en/dev/topics/auth/" rel="noreferrer">http://docs.djangoproject.com/en/dev/topics/auth/</a></p> <p>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.</p> <p>There are two parts you need to look at.</p> <ol> <li>Check that a user requesting a page has permission to do so.</li> <li>Only display links to the user if he has the permission.</li> </ol> <p>For 1. you can check permissions in a decorator as such:</p> <pre><code>from django.contrib.auth.decorators import permission_required @permission_required('polls.can_vote') def some_view(request): </code></pre> <p>For 2. the currently logged-in user's permissions are stored in the template variable {{ perms }}. This code checks the same permission as above.</p> <pre><code>{% if perms.polls.can_vote %} &lt;a href="/vote"&gt;vote&lt;/a&gt; {% endif %} </code></pre> <p>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:</p> <pre><code>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), []) </code></pre> <p>There are probably many other approaches.</p>
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    1. COHow would you generate a list of links to provide a user using django's built-in permissions system?
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    2. COFor the record, I did not select this answer as the best. I don't think it is.
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