We can chain querysets together with the chain(…) function ^[python-doc] of the itertools package ^[python-doc]. For example if we have two Posts, we can chain the posts with a publish_date and then the ones where the publish_date is NULL:

from itertools import chain

qs1 = Post.objects.filter(publish_date__isnull=False).order_by('publish_date')
qs2 = Post.objects.filter(publish_date=None)

result = chain(qs1, qs2)

Why is it a problem?

The main problem is that the result is not a QuerySet, but a chain object. This means that all methods offered by a QuerySet can no longer be used. Indeed, say that we want to filter the Posts with:

result.filter(author=some_author)

then this will raise an error. Often such filtering is not done explicitly in the view, but for example by a FilterSet the developer wants to use.

Another problem is that a chain can not be enumerated multiple times. Indeed:

>>> c = chain([1,4], [2,5])
>>> list(c)
[1, 4, 2, 5]
>>> list(c)
[]

This thus means if multiple for loops are used, only the first will iterate over the elements. We can work with list(…), and thus use result = list(chain(qs1, qs2)) to prevent this effect.

Another problem is that result will eventually perform multiple queries. In this example there will be two queries. If we chain however five querysets together, it results in (at least) five queries. This thus makes it more expensive.

What can be done to resolve the problem?

Group the queries together into a single queryset. If the order is of no importance, we can make use of the | operator:

result = qs1 | qs2

if the order is of importance, we can make use of .union(…) ^[Django-doc]:

qs1.union(qs2, all=True)

Extra tips

We can use chain(…) when we query for example different models like:

from itertools import chain

qs1 = Post.objects.all()
qs2 = Author.objects.all()

result = list(chain(qs1, qs2))

But it is seldomly the case that a collection contains elements of a different type. Especially since very often processing Posts will be different from processing Authors.