Simplifying flags when filtering models-CodePudding

I have a user model that has a set of notifications:

class User(AbstractUser):

    # Notification flags
    send_to_all = models.BooleanField(default=True)
    send_to_me = models.BooleanField(default=True)
    send_to_friends = models.BooleanField(default=True)
    # ...

I am passing them from front-end to my views where I trigger send_notifications function.

    # ...

    def update(self, request, *args, **kwargs):
        # ...
        send_notifications(request.data.get('send_to_all', False))

    # ...

And in the send_notifications I want to query User.objects.filter() and inlcude only those that have the flag passed from front-end set to True.

My question is:

Is there any way to simplify it? I am asking because there is many notifications options, like 20. And currently my code has 20 ifs:

def send_notifications(flag):
    if flag == 'send_to_all':
       users_to_send = User.objects.filter(send_to_all=True) 
    if flag == 'send_to_me':
       users_to_send = User.objects.filter(send_to_me=True) 
    if flag == 'send_to_friends':
       users_to_send = User.objects.filter(send_to_friends=True) 
    if flag == 'another_send_flag':
       users_to_send = User.objects.filter(another_send_flag=True) 
    if flag == 'another_send_flag':
       users_to_send = User.objects.filter(another_send_flag=True)

Thanks!

CodePudding user response：

You can wrap User.objects.filter() with eval like that:

users_to_send = eval(f"User.objects.filter({flag}=True)")