id	summary	reporter	owner	description	type	status	component	version	severity	resolution	keywords	cc	stage	has_patch	needs_docs	needs_tests	needs_better_patch	easy	ui_ux
33477	Union Operator merges annotated values wrongfully	Tobias Maschek	nobody	"Hi, maybe I'm doing something majorly incorrectly, but Django merges Querysets at least in an unexpected way.

= Setup
I have the following models:
{{{
class Employee(models.Model):
    name = models.CharField(max_length=64)

class Shift(models.Model):
    employee = models.ForeignKey(Employee, on_delete=models.CASCADE)
    start = models.DateTimeField()
    dur = models.DecimalField(max_digits=4, decimal_places=2, default=0)
}}}

= Querysets
With
{{{ employees_with_shifts = Employee.objects.filter(shift__start__range=(monday, monday + timedelta(weeks=1))).annotate(total=Sum('shift__dur')) }}}
I can get all employees, that have at least one shift in the week containing the given monday, with the annotated total worktime.


With
{{{
employees_without_shifts = Employee.objects.annotate(total=Value(0, IntegerField()))
    .exclude(shift__start__range=(monday, monday + timedelta(weeks=1)))
}}}
I get all employees without any shifts and on all objects is the total annotated with 0. So far, so good.

= The Problem

But if I'm now combining these two sets to `employees = employees_with_shifts | employees_without_shifts`, **the annotated total value is not any more zero** for all employees in `employees_without_shifts `, instead it's some sum of all shifts (ignoring the week). Even when the exclude statement is removed, it changes nothing. 

I suspect that Django combines these two Querysets before evaluating, and in the combination process the range gets thrown off.

If I'm combining these two sets with `employees = sorted(chain(employees_with, employees_without), key=lambda instance: instance.id)` the annotated value is correct.

Or do I use the union operator incorrectly?"	Bug	closed	Database layer (models, ORM)	4.0	Normal	invalid	annotation, querysets, merging	Tobias Maschek	Unreviewed	0	0	0	0	0	0