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copied!<p>Simply do:</p> <pre><code>def has_roles(user, roles): return user.profile.user_types.filter(pk__in=roles).count() == len(roles) print has_roles(user, [6,7,8]) </code></pre> <p>P.S. I would shy away from using hard coded PK ID numbers as your identifiers. Too many things can go wrong with that. Instead define a mapping at runtime and reference them by names. Multiple things will improve:</p> <ul> <li>Your code will be much easier to read </li> <li>Django will offload model instance matching to the DB the first item</li> <li>From then on you can cache the ID and not ask the DB again</li> </ul> <p>Then you can do:</p> <pre><code>class UserType( models.Model ): TYPES = (( 'tenant', 'Tenant'), ( 'propman', 'Property Manager'), ( 'propown', 'Property Owner'), ( 'vendman', 'Vendor Manager'), ( 'vendor', 'Vendor'), ( 'viewer', 'Viewer'), ( 'moderator', 'Moderator'), ( 'admin', 'Administrator')) name = models.CharField( max_length = 135, choices=TYPES ) def has_role(user, role): return user.profile.user_types.filter(name=role).count() == 1 def has_roles(user, roles): return user.profile.user_types.filter(name__in=roles).count() == len(roles) print has_roles(user, ['viewer','moderator','admin']) </code></pre> <p>Finally, you can add the two functions above to:</p> <pre><code>class UserProfile( models.Model ) : user = models.OneToOneField( User ) user_types = models.ManyToManyField( UserType, null = True, blank = True ) def has_role(self, role): return self.user_types.filter(name=role).count() == 1 def has_roles(self, roles): return self.user_types.filter(name__in=roles).count() == len(roles) </code></pre> <p>And then use it like this in the future:</p> <pre><code>u = User.objects.get(username='me') if u.userprofile.has_role('admin'): print 'I have the powah!' </code></pre>

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