Elegant way of fetching multiple objects in custom order - django

Question

What's an elegant way for fetching multiple objects in some custom order from a DB in django?
For example, suppose you have a few products, each with its name, and you want to fetch three of them to display in a row on your website page, in some fixed custom order. Suppose the names of the products which you want to display are, in order: ["Milk", "Chocolate", "Juice"]
One could do
unordered_products = Product.objects.filter(name__in=["Milk", "Chocolate", "Juice"])
products = [
unordered_products.filter(name="Milk")[0],
unordered_products.filter(name="Chocolate")[0],
unordered_products.filter(name="Juice")[0],
]
And the post-fetch ordering part could be improved to use a name-indexed dictionary instead:
ordered_product_names = ["Milk", "Chocolate", "Juice"]
products_by_name = dict((x.name, x) for x in unordered_products)
products = [products_by_name[name] for name in ordered_product_names]
But is there a more elegant way? e.g., convey the desired order to the DB layer somehow, or return the products grouped by their name (aggregation seems to be similar to what I want, but I want the actual objects, not statistics about them).

Answer 1

You can order your product by a custom order with only one query of your ORM (executing one SQL query only):
ordered_products = Product.objects.filter(
name__in=['Milk', 'Chocolate', 'Juice']
).annotate(
order=Case(
When(name='Milk', then=Value(0)),
When(name='Chocolate', then=Value(1)),
When(name='Juice', then=Value(2)),
output_field=IntegerField(),
)
).order_by('order')
Update
Note
Speaking about "elegant way" (and best practice) I think extra method (proposed by #Satendra) is absolutely to avoid.
Official Django documentation report this about extra :
Warning
You should be very careful whenever you use extra(). Every time you
use it, you should escape any parameters that the user can control by
using params in order to protect against SQL injection attacks .
Please read more about SQL injection protection.
Optimized version
If you want to handle more items whit only one query you can change my first query and use the Django ORM flexibility as suggested by #Shubhanshu in his answer:
products = ['Milk', 'Chocolate', 'Juice']
ordered_products = Product.objects.filter(
name__in=products
).order_by(Case(
*[When(name=n, then=i) for i, n in enumerate(products)],
output_field=IntegerField(),
))
The output of this command will be similar to this:
<QuerySet [<Product: Milk >, <Product: Chocolate>, <Product: Juice>]>
And the SQL generated by the ORM will be like this:
SELECT "id", "name"
FROM "products"
WHERE "name" IN ('Milk', 'Chocolate', 'Juice')
ORDER BY CASE
WHEN "name" = 'Milk' THEN 0
WHEN "name" = 'Chocolate' THEN 1
WHEN "name" = 'Juice' THEN 2
ELSE NULL
END ASC

Answer 2

When there is no relation between the objects that you are fetching and you still wish to fetch (or arrange) them in certain (custom) order, you may try doing this:
unordered_products = Product.objects.filter(name__in=["Milk", "Chocolate", "Juice"])
product_order = ["Milk", "Chocolate", "Juice"]
preserved = Case(*[When(name=name, then=pos) for pos, name in enumerate(product_order)])
ordered_products = unordered_products.order_by(preserved)
Hope it helps!

Answer 3

Try this into meta class from model:
class Meta:
ordering = ('name', 'related__name', )
this get your records ordered by your specified field's
then: chocolate, chocolate blue, chocolate white, juice green, juice XXX, milk, milky, milk YYYY should keep that order when you fetch

Answer 4

Creating a QuerySet from a list while preserving order
This means the order of output QuerySet will be same as the order of list used to filter it.
The solution is more or less same as #PaoloMelchiorre answer
But if there are more items lets say 1000 products in
product_names then you don't have to worry about adding more conditions in Case, you can use extra method of QuerySet
product_names = ["Milk", "Chocolate", "Juice", ...]
clauses = ' '.join(['WHEN name=%s THEN %s' % (name, i) for i, name in enumerate(product_names)])
ordering = 'CASE %s END' % clauses
queryset = Product.objects.filter(name__in=product_names).extra(
select={'ordering': ordering}, order_by=('ordering',))
# Output: <QuerySet [<Product: Milk >, <Product: Chocolate>, <Product: Juice>,...]>