模板中的Django过滤器related_name子集

Question

I have a foreign key and I'm using the related_name field like so:

class Pizza(models.Model):
   ...
   restaurant = models.ForeignKey('Restaurant', related_name='pizzas_offered')
   active = models.BooleanField(...)
   ...

Example from the view:

my_restaurant = get_object_or_404(Restaurant, pk=id)

In any template I can run something like my_restaurant.pizzas_offered.all to get all the pizzas belonging to a particular restaurant. However, I only want the pizzas that are active (active=True). Is there a way to retrieve this subset in the template, without having to pass a separate variable in the view? Please note that I always want to only show the active pizzas only so if I have to make a change in the model to make this happen, that is fine too.

NOTE: in the view I can simply pass my_restaurant.pizzas_offered.filter(active=True) but it returns an error when I use it in the template:

{% for details in my_restaurant.pizzas_offered.filter(active=True) %}
  {{ details.name }}
{% endfor %}

It returns this error: Could not parse the remainder: '(active=True)'

There are some reasons why I want to do this on template level and not in the view (main reason: I often loop through all the records in the database and not just one restaurant, so I can't just query for the one record). So my question is how to do this on template-level.

Answer 1

You need to create a Manager for your Pizza model and set the Meta.base_manager_name :

class PizzaManager(models.Manager):
    def active(self):
        return self.filter(status=True)

class Pizza(models.Model):
    ...
    objects = PizzaManager()

    class meta:
        base_manager_name = 'objects'

Now you can use the method active in your template:

{% for details in my_restaurant.pizzas_offered.active %}
    ...
{% endfor %}

For more information you can read the documentation about Default and Base managers .