Custom QuerySet and Manager without breaking DRY?

DjangoDjango ModelsDjango QuerysetDjango Managers

Django Problem Overview


I'm trying to find a way to implement both a custom QuerySet and a custom Manager without breaking DRY. This is what I have so far:

class MyInquiryManager(models.Manager):
    def for_user(self, user):
        return self.get_query_set().filter(
                    Q(assigned_to_user=user) |
                    Q(assigned_to_group__in=user.groups.all())
                )

class Inquiry(models.Model):   
    ts = models.DateTimeField(auto_now_add=True)
    status = models.ForeignKey(InquiryStatus)
    assigned_to_user = models.ForeignKey(User, blank=True, null=True)
    assigned_to_group = models.ForeignKey(Group, blank=True, null=True)
    objects = MyInquiryManager()

This works fine, until I do something like this:

inquiries = Inquiry.objects.filter(status=some_status)
my_inquiry_count = inquiries.for_user(request.user).count()

This promptly breaks everything because the QuerySet doesn't have the same methods as the Manager. I've tried creating a custom QuerySet class, and implementing it in MyInquiryManager, but I end up replicating all of my method definitions.

I also found this snippet which works, but I need to pass in the extra argument to for_user so it breaks down because it relies heavily on redefining get_query_set.

Is there a way to do this without redefining all of my methods in both the QuerySet and the Manager subclasses?

Django Solutions


Solution 1 - Django

The Django 1.7 released a new and simple way to create combined queryset and model manager:

class InquiryQuerySet(models.QuerySet):
    def for_user(self, user):
        return self.filter(
            Q(assigned_to_user=user) |
            Q(assigned_to_group__in=user.groups.all())
        )

class Inquiry(models.Model):
    objects = InqueryQuerySet.as_manager()

See Creating Manager with QuerySet methods for more details.

Solution 2 - Django

Django has changed! Before using the code in this answer, which was written in 2009, be sure to check out the rest of the answers and the Django documentation to see if there is a more appropriate solution.


The way I've implemented this is by adding the actual get_active_for_account as a method of a custom QuerySet. Then, to make it work off the manager, you can simply trap the __getattr__ and return it accordingly

To make this pattern re-usable, I've extracted out the Manager bits to a separate model manager:

custom_queryset/models.py

from django.db import models
from django.db.models.query import QuerySet

class CustomQuerySetManager(models.Manager):
    """A re-usable Manager to access a custom QuerySet"""
    def __getattr__(self, attr, *args):
        try:
            return getattr(self.__class__, attr, *args)
        except AttributeError:
            # don't delegate internal methods to the queryset
            if attr.startswith('__') and attr.endswith('__'):
                raise
            return getattr(self.get_query_set(), attr, *args)

    def get_query_set(self):
        return self.model.QuerySet(self.model, using=self._db)

Once you've got that, on your models all you need to do is define a QuerySet as a custom inner class and set the manager to your custom manager:

your_app/models.py

from custom_queryset.models import CustomQuerySetManager
from django.db.models.query import QuerySet

class Inquiry(models.Model):
    objects = CustomQuerySetManager()

    class QuerySet(QuerySet):
        def active_for_account(self, account, *args, **kwargs):
            return self.filter(account=account, deleted=False, *args, **kwargs)

With this pattern, any of these will work:

>>> Inquiry.objects.active_for_account(user)
>>> Inquiry.objects.all().active_for_account(user)
>>> Inquiry.objects.filter(first_name='John').active_for_account(user)

UPD if you are using it with custom user(AbstractUser), you need to change
from

class CustomQuerySetManager(models.Manager):

to

from django.contrib.auth.models import UserManager

class CustomQuerySetManager(UserManager):
    ***

Solution 3 - Django

You can provide the methods on the manager and queryset using a mixin.

This also avoids the use of a __getattr__() approach.

from django.db.models.query import QuerySet

class PostMixin(object):
    def by_author(self, user):
        return self.filter(user=user)

    def published(self):
        return self.filter(published__lte=datetime.now())

class PostQuerySet(QuerySet, PostMixin):
    pass

class PostManager(models.Manager, PostMixin):
    def get_query_set(self):
        return PostQuerySet(self.model, using=self._db)

Solution 4 - Django

You can now use the from_queryset() method on you manager to change its base Queryset.

This allows you to define your Queryset methods and your manager methods only once

from the docs

> For advanced usage you might want both a custom Manager and a custom QuerySet. You can do that by calling Manager.from_queryset() which returns a subclass of your base Manager with a copy of the custom QuerySet methods:

class InqueryQueryset(models.Queryset):
    def custom_method(self):
        """ available on all default querysets"""

class BaseMyInquiryManager(models.Manager):
    def for_user(self, user):
        return self.get_query_set().filter(
                    Q(assigned_to_user=user) |
                    Q(assigned_to_group__in=user.groups.all())
                )

MyInquiryManager = BaseInquiryManager.from_queryset(InquiryQueryset)

class Inquiry(models.Model):   
    ts = models.DateTimeField(auto_now_add=True)
    status = models.ForeignKey(InquiryStatus)
    assigned_to_user = models.ForeignKey(User, blank=True, null=True)
    assigned_to_group = models.ForeignKey(Group, blank=True, null=True)
    objects = MyInquiryManager()

Solution 5 - Django

A slightly improved version of T. Stone’s approach:

def objects_extra(mixin_class):
    class MixinManager(models.Manager, mixin_class):
        class MixinQuerySet(QuerySet, mixin_class):
            pass

        def get_query_set(self):
            return self.MixinQuerySet(self.model, using=self._db)

    return MixinManager()

Class decorators make usage as simple as:

class SomeModel(models.Model):
    ...
    @objects_extra
    class objects:
        def filter_by_something_complex(self, whatever parameters):
            return self.extra(...)
        ...

Update: support for nonstandard Manager and QuerySet base classes, e. g. @objects_extra(django.contrib.gis.db.models.GeoManager, django.contrib.gis.db.models.query.GeoQuerySet):

def objects_extra(Manager=django.db.models.Manager, QuerySet=django.db.models.query.QuerySet):
    def oe_inner(Mixin, Manager=django.db.models.Manager, QuerySet=django.db.models.query.QuerySet):
        class MixinManager(Manager, Mixin):
            class MixinQuerySet(QuerySet, Mixin):
                pass

            def get_query_set(self):
                return self.MixinQuerySet(self.model, using=self._db)

        return MixinManager()

    if issubclass(Manager, django.db.models.Manager):
        return lambda Mixin: oe_inner(Mixin, Manager, QuerySet)
    else:
        return oe_inner(Mixin=Manager)

Solution 6 - Django

There are use-cases where we need to call custom QuerySet methods from the manager instead of using the get_manager method of a QuerySet.

A mixin would suffice based on the solution posted in one of the accepted solution comments.

class CustomQuerySetManagerMixin:
    """
    Allow Manager which uses custom queryset to access queryset methods directly.
    """
    def __getattr__(self, name):
        # don't delegate internal methods to queryset
        # NOTE: without this, Manager._copy_to_model will end up calling
        # __getstate__ on the *queryset* which causes the qs (as `all()`)
        #  to evaluate itself as if it was being pickled (`len(self)`)
        if name.startswith('__'):
            raise AttributeError
        return getattr(self.get_queryset(), name)

For example,

class BookQuerySet(models.QuerySet):
    def published(self):
        return self.filter(published=True)

    def fiction(self):
        return self.filter(genre="fiction")

    def non_fiction(self):
        return self.filter(genre="non-fiction")

class BookManager(CustomQuerySetManagerMixin, models.Manager):
    def get_queryset(self):
        return BookQuerySet(self.model, using=self._db).published()

class Book(models.Model):
    title = models.CharField(max_length=200)
    genre = models.CharField(choices=[('fiction', _('Fiction')), ('non-fiction', _('Non-Fiction'))])
    published = models.BooleanField(default=False)
    author = models.ForeignKey(Author, on_delete=models.CASCADE, related_name="books")

    objects = BookManager()

class Author(models.Model):
    name = models.CharField(max_length=200)

With the above, we can access related objects (Book) like below without defining new methods in the manager for each queryset method.

fiction_books = author.books.fiction()

Solution 7 - Django

based on django 3.1.3 source code, i found a simple solution

from django.db.models.manager import BaseManager

class MyQuerySet(models.query.QuerySet):
      def my_custom_query(self):
          return self.filter(...)

class MyManager(BaseManager.from_queryset(MyQuerySet)):
     ...

class MyModel(models.Model):
     objects = MyManager()

Solution 8 - Django

The following works for me.

def get_active_for_account(self,account,*args,**kwargs):
    """Returns a queryset that is 
    Not deleted
    For the specified account
    """
    return self.filter(account = account,deleted=False,*args,**kwargs)

This is on the default manager; so I used to do something like:

Model.objects.get_active_for_account(account).filter()

But there is no reason it should not work for a secondary manager.

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJack M.View Question on Stackoverflow
Solution 1 - DjangoiMom0View Answer on Stackoverflow
Solution 2 - DjangoT. StoneView Answer on Stackoverflow
Solution 3 - DjangovdboorView Answer on Stackoverflow
Solution 4 - DjangomaazzaView Answer on Stackoverflow
Solution 5 - DjangoRoman OdaiskyView Answer on Stackoverflow
Solution 6 - DjangoSian Lerk LauView Answer on Stackoverflow
Solution 7 - DjangoEbrahim KarimiView Answer on Stackoverflow
Solution 8 - DjangolprsdView Answer on Stackoverflow