Can I use a single search bar for searching different things? - django

Question

I'm trying to use a single search bar for searching different models. Is it possible to handle it with one view?
my views.py file:
class SearchResultsView(ListView):
model = Food
template_name = 'restaurant/food_search_results.html'
context_object_name = 'data'
def get_queryset(self):
query = self.request.GET.get('search')
if query is not None:
return Food.objects.filter(name__icontains=query)
else:
return Food.objects.none()
my models.py file:
class Food(models.Model):
name = models.CharField(max_length=100)
image = models.ImageField(upload_to=upload_path)
description = models.TextField(max_length=500)
created = models.DateTimeField(auto_now_add=True)
meal_category = models.ManyToManyField(MealCategory, related_name="meal")
food_restaurant_category = models.ManyToManyField(FoodRestaurantCategory, related_name="food_cat")
class Restaurant(models.Model):
name = models.CharField(max_length=100)

Answer 1

You can use the context to pass the second model filtered list:
class SearchResultsView(ListView):
model = Food
template_name = 'restaurant/food_search_results.html'
context_object_name = 'data'
def get_queryset(self):
query = self.request.GET.get('search')
if query is not None:
return Food.objects.filter(name__icontains=query)
else:
return Food.objects.none()
def get_context_data(self, **kwargs):
query = self.request.GET.get('search')
context = super(SearchResultsView, self).get_context_data(**kwargs)
if query is not None:
filtered_restaurants= Restaurant.objects.filter(name__icontains=query)
else:
filtered_restaurants= Restaurant.objects.none()
context.update({
'restaurants_list': filtered_restaurants
})
return context