Re: How to access using foreing key

If you are looking for all of the Restaurants with a 5.0 rating, it looks like

Restaurant.objects.filter(rating_restaurant__rating = 5.0)

If you are looking for Restaurants with an average rating of 5.0, that's a bit trickier.

from django.db.models import Avg

Restaurant.objects.annotate(avg_rating = Avg('rating_restaurant__rating')).filter(avg_rating__eq = 5.0)

You can find some surprisingly relevant examples at https://docs.djangoproject.com/en/2.2/topics/db/aggregation/#order-of-annotate-and-filter-clauses.

On Oct 30, 2020, at 9:52 AM, Ashutosh Mishra <ashutoshmishra333@gmail.com> wrote:

class User(models.Model):

name=models.CharField(max_length=50,blank=True,null=True)

latitude = models.CharField('Latitude', max_length=30, blank=True, null=True)

longitude = models.CharField('Longitude', max_length=30, blank=True, null=True)

location = models.PointField(blank=True, null=True)

def __str__(self):

return str(self.name)

class Restaurant(models.Model):

restaurant_name=models.CharField(max_length=50,blank=True,null=True)

latitude = models.CharField('Latitude', max_length=30, blank=True, null=True)

longitude = models.CharField('Longitude', max_length=30, blank=True, null=True)

location = models.PointField(blank=True, null=True)

def __str__(self):

return str(self.restaurant_name)

class RestaurantRating(models.Model):

user = models.ForeignKey(User,on_delete=models.CASCADE,null=True,related_name="customer_rating")

restaurant = models.ForeignKey(Restaurant, on_delete=models.CASCADE, null=True, related_name="rating_restaurant")

rating = models.FloatField()

def __str__(self):

return str(self.user)

How can i filter the restaurants whose rating is 5,Restaurant.objects.filter()

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