django distinct query using custom equivalence

Say that my model looks like this:

class Alert(models.Model):
    datetime_alert = models.DateTimeField()
    alert_type = models.ForeignKey(Alert_Type, on_delete=models.CASCADE)
    dismissed = models.BooleanField(default=False)
    datetime_dismissed = models.DateTimeField(null=True)
    auid = models.CharField(max_length=64, unique=True)
    entities = models.ManyToManyField(to='Entity', through='Entity_To_Alert_Map')
    objects = Alert_Manager()

    def __eq__(self, other):
        return isinstance(other,
                          self.__class__) and self.alert_type == other.alert_type and \
               self.entities.all() == other.entities().all() and self.dismissed == other.dismissed

    def __ne__(self, other):
        return not self.__eq(other)

what I'm trying to accomplish is say this: two alert objects are equivalent if the dismissed status, alert type, and the associated entities are the same. Using this idea, is it possible to write a query to ask for all the distinct alerts based off that criteria? Selecting all of them and then filtering them out doesn't seem appealing.

Solution

You mention one method to do it, and I don't think it is very bad. I'm not aware of anything in Django that can do this.

However, I want you to think why this problem arises? If two alerts are equal if message, status and type is the same, then maybe this should be it's own class. I would consider creating another class DistinctAlert (or some better name) and have a foreign key to this class from Alert. Or even better, have one class that is Alert, and one that is called AlertEvent(your Alert class).

Would this solve your problem?

Edit:

Actually, there is a way to do this. You can combine values() and distinct(). This way, your query will be

Alert.objects.all().values("alert_type", "dismissed", "entities").distinct()

This will return a dictionary.

See more in the documentation of values()