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
<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 <ashutoshmishra...@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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