[Django]-Django app for user/account settings

3đź‘Ť

âś…

You’re probably going to need to create your own system for this. It shouldn’t be very difficult.

class Setting(models.Model):
    name = models.CharField(..)
    value = models.CharField(..)
    users = models.ManyToManyField(User)
    groups = models.ManyToManyField(Group)

    @classmethod
    def has_setting(cls, name, user):
        user_settings = cls.objects.filter(name=name, users__in=[user]).count()
        group_settings = cls.objects.filter(name=name, groups__in=user.groups.all()).count()
        return user_settings or group_settings

    @classmethod
    def get_settings(cls, name, user, priority=User):
        user_settings = list(cls.objects.filter(name=name, users__in=[user]))
        group_settings = list(cls.objects.filter(name=name, groups__in=user.groups.all()))
        if user_settings and not group_settings: 
            return user_settings
        elif group_settings and not user_settings:
            return group_settings
        else:
            return (group_settings, user_settings)[priority==User]

Then you can do something like this in your views:

def display_frob(request):
    if Setting.has_setting('display_frob', request.user):
         settings = Setting.get_setting('display_from', request.user, priority=Group)
         values = [setting.value for setting in settings]
         # if the User was in many groups, we now have a list of values, for the setting
         # `display_frob`, that aligns with each of those groups

You can easy build a decorator which will do the check, and provide a list (or a single item) of values to the view.

👤Josh Smeaton

0đź‘Ť

For permissions for “actions” (where an action is typically implemented by a view) I commonly use decorators.

The user_passes_test decorator is great for this kind of purpose.

You could create user permissions not linked to models.

0đź‘Ť

Maybe this is what you looking for : django-guardian

👤Aldarund

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