Queryset bug when using Q()? -- doubles up the counts by doing an extra join.

[Javier Buzzi]

I ran this on all the "recent" versions of python and i see the issue on all of them. At this point, im not 100% sure if it is a problem or not, i ran across this trying to show some counts in the Admin that had some search_fields that traversed the model backwards and forwards in order to match the thing properly.

This "bug" can be seen from 1.11-3.0. Haven't checked 3.1 because i cant find it in pip.

This is my mocked models of the issue:

class ModelA(models.Model):
    somecriteria = models.CharField(max_length=50)


class ModelC(models.Model):
    unimportant = models.CharField(max_length=50)


class ModelB(models.Model):
    somerelation = models.ForeignKey(ModelA, on_delete=models.CASCADE)
    m2m = models.ManyToManyField(
        ModelC,
        through='ModelBC'
    )


class ModelBC(models.Model):
    b = models.ForeignKey(ModelB, on_delete=models.CASCADE)
    c = models.ForeignKey(ModelC, on_delete=models.CASCADE)

The "bug" is shown when i do:

queryset = ModelB.objects.select_related('a').annotate(num_c=Count('m2m__id')).values('num_c')
queryset = queryset.filter(Q(modelbc__c__unimportant='1') | Q(somerelation__somecriteria='1'))

When i do:

queryset = ModelB.objects.select_related('a').annotate(num_c=Count('m2m__id')).values('num_c')
queryset = queryset.filter(modelbc__c__unimportant='1', somerelation__somecriteria='1')

it works as expected.

My testing arena can be found here: ​https://gist.github.com/kingbuzzman/9fe5470e31a421aa88b2a64e5447e147 see the comments, for the way to run it.

[Mariusz Felisiak]

These querysets are different so you cannot expect the same results:

• Q(condition1) | Q(condition2) means "condition1 OR condition2",
• .filter(condition1, condition2) means "condition1 AND condition2".

[Javier Buzzi]

My apologies, i just realized my mistake. Looks like im going to have to wrap the admin filters in order for it to work correctly, and thats fine. The ORM is working as expected. Sorry to waste anyones time.