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Home/ Questions/Q 351463
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T11:38:46+00:00 2026-05-12T11:38:46+00:00

Given the Model: class Profile(models.Model): user = models.ForeignKey(User, unique=True) class Thingie(models.Model): children = models.ManyToManyField(‘self’,

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Given the Model:

class Profile(models.Model):
    user = models.ForeignKey(User, unique=True)

class Thingie(models.Model):
    children = models.ManyToManyField('self', blank=True, symmetrical=False) 

class Relation(models.Model):
    profile = models.ForeignKey(Profile)
    thingie = models.ForeignKey(Thingie)

How would one return a QuerySet containing all Profile instances related to a given Thingie? That is, every Profile that has a foreign key pointing to it from a Relation and to the given thingie.

I know all about select_related(), and how I could use it to do this by iterating but i find iterating irritating (badoop bah!). Also, values_list() has been looked at, but it doesn’t quite do the right thing.

Please help! Thanks!

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  1. Editorial Team
    Editorial Team
    2026-05-12T11:38:46+00:00Added an answer on May 12, 2026 at 11:38 am

    Do you definitely need it to be a queryset? If you only need it to be an iterable, a simple expression for your purposes is:

    profiles = [r.profile for r in thingie.relation_set.all()]
    

    I’m not sure if a list comprehension counts as irritating iterating, but to me this is a perfectly intuitive, pythonic approach. Of course if you need it to be a queryset, you’re going to do something messier with two queries, eg:

    relation_values = thingie.relation_set.all().values_list('pk', flat=True)
    profiles = Profile.objects.filter(relation__in=relation_values)
    

    See the “in” documentation for more. I prefer the first approach, if you don’t need a queryset. Oh and if you only want distinct profiles you can just take set(profiles) in the first approach or use the distinct() queryset method in the second approach.

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