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This patch does not remove all occurrences of the words in question. Rather, I went through all of the occurrences of the words listed below, and judged if they a) suggested the reader had some kind of knowledge/experience, and b) if they added anything of value (including tone of voice, etc). I left most of the words alone. I looked at the following words: - simply/simple - easy/easier/easiest - obvious - just - merely - straightforward - ridiculous Thanks to Carlton Gibson for guidance on how to approach this issue, and to Tim Bell for providing the idea. But the enormous lion's share of thanks go to Adam Johnson for his patient and helpful review.
456 lines
17 KiB
Plaintext
456 lines
17 KiB
Plaintext
========
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Managers
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========
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.. currentmodule:: django.db.models
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.. class:: Manager()
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A ``Manager`` is the interface through which database query operations are
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provided to Django models. At least one ``Manager`` exists for every model in
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a Django application.
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The way ``Manager`` classes work is documented in :doc:`/topics/db/queries`;
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this document specifically touches on model options that customize ``Manager``
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behavior.
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.. _manager-names:
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Manager names
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=============
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By default, Django adds a ``Manager`` with the name ``objects`` to every Django
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model class. However, if you want to use ``objects`` as a field name, or if you
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want to use a name other than ``objects`` for the ``Manager``, you can rename
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it on a per-model basis. To rename the ``Manager`` for a given class, define a
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class attribute of type ``models.Manager()`` on that model. For example::
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from django.db import models
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class Person(models.Model):
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#...
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people = models.Manager()
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Using this example model, ``Person.objects`` will generate an
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``AttributeError`` exception, but ``Person.people.all()`` will provide a list
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of all ``Person`` objects.
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.. _custom-managers:
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Custom managers
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===============
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You can use a custom ``Manager`` in a particular model by extending the base
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``Manager`` class and instantiating your custom ``Manager`` in your model.
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There are two reasons you might want to customize a ``Manager``: to add extra
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``Manager`` methods, and/or to modify the initial ``QuerySet`` the ``Manager``
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returns.
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Adding extra manager methods
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----------------------------
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Adding extra ``Manager`` methods is the preferred way to add "table-level"
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functionality to your models. (For "row-level" functionality -- i.e., functions
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that act on a single instance of a model object -- use :ref:`Model methods
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<model-methods>`, not custom ``Manager`` methods.)
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A custom ``Manager`` method can return anything you want. It doesn't have to
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return a ``QuerySet``.
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For example, this custom ``Manager`` offers a method ``with_counts()``, which
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returns a list of all ``OpinionPoll`` objects, each with an extra
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``num_responses`` attribute that is the result of an aggregate query::
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from django.db import models
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class PollManager(models.Manager):
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def with_counts(self):
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from django.db import connection
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with connection.cursor() as cursor:
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cursor.execute("""
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SELECT p.id, p.question, p.poll_date, COUNT(*)
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FROM polls_opinionpoll p, polls_response r
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WHERE p.id = r.poll_id
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GROUP BY p.id, p.question, p.poll_date
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ORDER BY p.poll_date DESC""")
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result_list = []
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for row in cursor.fetchall():
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p = self.model(id=row[0], question=row[1], poll_date=row[2])
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p.num_responses = row[3]
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result_list.append(p)
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return result_list
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class OpinionPoll(models.Model):
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question = models.CharField(max_length=200)
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poll_date = models.DateField()
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objects = PollManager()
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class Response(models.Model):
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poll = models.ForeignKey(OpinionPoll, on_delete=models.CASCADE)
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person_name = models.CharField(max_length=50)
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response = models.TextField()
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With this example, you'd use ``OpinionPoll.objects.with_counts()`` to return
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that list of ``OpinionPoll`` objects with ``num_responses`` attributes.
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Another thing to note about this example is that ``Manager`` methods can
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access ``self.model`` to get the model class to which they're attached.
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Modifying a manager's initial ``QuerySet``
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------------------------------------------
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A ``Manager``’s base ``QuerySet`` returns all objects in the system. For
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example, using this model::
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from django.db import models
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class Book(models.Model):
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title = models.CharField(max_length=100)
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author = models.CharField(max_length=50)
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...the statement ``Book.objects.all()`` will return all books in the database.
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You can override a ``Manager``’s base ``QuerySet`` by overriding the
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``Manager.get_queryset()`` method. ``get_queryset()`` should return a
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``QuerySet`` with the properties you require.
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For example, the following model has *two* ``Manager``\s -- one that returns
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all objects, and one that returns only the books by Roald Dahl::
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# First, define the Manager subclass.
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class DahlBookManager(models.Manager):
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def get_queryset(self):
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return super().get_queryset().filter(author='Roald Dahl')
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# Then hook it into the Book model explicitly.
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class Book(models.Model):
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title = models.CharField(max_length=100)
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author = models.CharField(max_length=50)
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objects = models.Manager() # The default manager.
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dahl_objects = DahlBookManager() # The Dahl-specific manager.
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With this sample model, ``Book.objects.all()`` will return all books in the
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database, but ``Book.dahl_objects.all()`` will only return the ones written by
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Roald Dahl.
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Of course, because ``get_queryset()`` returns a ``QuerySet`` object, you can
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use ``filter()``, ``exclude()`` and all the other ``QuerySet`` methods on it.
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So these statements are all legal::
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Book.dahl_objects.all()
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Book.dahl_objects.filter(title='Matilda')
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Book.dahl_objects.count()
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This example also pointed out another interesting technique: using multiple
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managers on the same model. You can attach as many ``Manager()`` instances to
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a model as you'd like. This is a non-repetitive way to define common "filters"
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for your models.
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For example::
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class AuthorManager(models.Manager):
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def get_queryset(self):
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return super().get_queryset().filter(role='A')
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class EditorManager(models.Manager):
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def get_queryset(self):
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return super().get_queryset().filter(role='E')
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class Person(models.Model):
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first_name = models.CharField(max_length=50)
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last_name = models.CharField(max_length=50)
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role = models.CharField(max_length=1, choices=[('A', _('Author')), ('E', _('Editor'))])
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people = models.Manager()
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authors = AuthorManager()
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editors = EditorManager()
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This example allows you to request ``Person.authors.all()``, ``Person.editors.all()``,
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and ``Person.people.all()``, yielding predictable results.
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.. _default-managers:
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Default managers
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----------------
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.. attribute:: Model._default_manager
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If you use custom ``Manager`` objects, take note that the first ``Manager``
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Django encounters (in the order in which they're defined in the model) has a
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special status. Django interprets the first ``Manager`` defined in a class as
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the "default" ``Manager``, and several parts of Django (including
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:djadmin:`dumpdata`) will use that ``Manager`` exclusively for that model. As a
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result, it's a good idea to be careful in your choice of default manager in
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order to avoid a situation where overriding ``get_queryset()`` results in an
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inability to retrieve objects you'd like to work with.
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You can specify a custom default manager using :attr:`Meta.default_manager_name
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<django.db.models.Options.default_manager_name>`.
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If you're writing some code that must handle an unknown model, for example, in
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a third-party app that implements a generic view, use this manager (or
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:attr:`~Model._base_manager`) rather than assuming the model has an ``objects``
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manager.
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Base managers
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-------------
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.. attribute:: Model._base_manager
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.. _managers-for-related-objects:
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Using managers for related object access
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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By default, Django uses an instance of the ``Model._base_manager`` manager
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class when accessing related objects (i.e. ``choice.question``), not the
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``_default_manager`` on the related object. This is because Django needs to be
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able to retrieve the related object, even if it would otherwise be filtered out
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(and hence be inaccessible) by the default manager.
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If the normal base manager class (:class:`django.db.models.Manager`) isn't
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appropriate for your circumstances, you can tell Django which class to use by
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setting :attr:`Meta.base_manager_name
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<django.db.models.Options.base_manager_name>`.
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Base managers aren't used when querying on related models. For example, if the
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``Question`` model :ref:`from the tutorial <creating-models>` had a ``deleted``
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field and a base manager that filters out instances with ``deleted=True``, a
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queryset like ``Choice.objects.filter(question__name__startswith='What')``
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would include choices related to deleted questions.
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Don't filter away any results in this type of manager subclass
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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This manager is used to access objects that are related to from some other
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model. In those situations, Django has to be able to see all the objects for
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the model it is fetching, so that *anything* which is referred to can be
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retrieved.
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If you override the ``get_queryset()`` method and filter out any rows, Django
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will return incorrect results. Don't do that. A manager that filters results
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in ``get_queryset()`` is not appropriate for use as a base manager.
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.. _calling-custom-queryset-methods-from-manager:
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Calling custom ``QuerySet`` methods from the manager
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----------------------------------------------------
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While most methods from the standard ``QuerySet`` are accessible directly from
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the ``Manager``, this is only the case for the extra methods defined on a
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custom ``QuerySet`` if you also implement them on the ``Manager``::
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class PersonQuerySet(models.QuerySet):
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def authors(self):
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return self.filter(role='A')
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def editors(self):
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return self.filter(role='E')
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class PersonManager(models.Manager):
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def get_queryset(self):
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return PersonQuerySet(self.model, using=self._db)
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def authors(self):
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return self.get_queryset().authors()
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def editors(self):
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return self.get_queryset().editors()
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class Person(models.Model):
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first_name = models.CharField(max_length=50)
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last_name = models.CharField(max_length=50)
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role = models.CharField(max_length=1, choices=[('A', _('Author')), ('E', _('Editor'))])
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people = PersonManager()
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This example allows you to call both ``authors()`` and ``editors()`` directly from
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the manager ``Person.people``.
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.. _create-manager-with-queryset-methods:
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Creating a manager with ``QuerySet`` methods
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--------------------------------------------
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In lieu of the above approach which requires duplicating methods on both the
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``QuerySet`` and the ``Manager``, :meth:`QuerySet.as_manager()
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<django.db.models.query.QuerySet.as_manager>` can be used to create an instance
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of ``Manager`` with a copy of a custom ``QuerySet``’s methods::
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class Person(models.Model):
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...
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people = PersonQuerySet.as_manager()
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The ``Manager`` instance created by :meth:`QuerySet.as_manager()
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<django.db.models.query.QuerySet.as_manager>` will be virtually
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identical to the ``PersonManager`` from the previous example.
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Not every ``QuerySet`` method makes sense at the ``Manager`` level; for
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instance we intentionally prevent the :meth:`QuerySet.delete()
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<django.db.models.query.QuerySet.delete>` method from being copied onto
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the ``Manager`` class.
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Methods are copied according to the following rules:
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- Public methods are copied by default.
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- Private methods (starting with an underscore) are not copied by default.
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- Methods with a ``queryset_only`` attribute set to ``False`` are always copied.
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- Methods with a ``queryset_only`` attribute set to ``True`` are never copied.
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For example::
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class CustomQuerySet(models.QuerySet):
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# Available on both Manager and QuerySet.
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def public_method(self):
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return
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# Available only on QuerySet.
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def _private_method(self):
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return
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# Available only on QuerySet.
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def opted_out_public_method(self):
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return
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opted_out_public_method.queryset_only = True
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# Available on both Manager and QuerySet.
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def _opted_in_private_method(self):
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return
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_opted_in_private_method.queryset_only = False
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``from_queryset()``
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~~~~~~~~~~~~~~~~~~~
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.. classmethod:: from_queryset(queryset_class)
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For advanced usage you might want both a custom ``Manager`` and a custom
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``QuerySet``. You can do that by calling ``Manager.from_queryset()`` which
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returns a *subclass* of your base ``Manager`` with a copy of the custom
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``QuerySet`` methods::
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class BaseManager(models.Manager):
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def manager_only_method(self):
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return
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class CustomQuerySet(models.QuerySet):
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def manager_and_queryset_method(self):
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return
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class MyModel(models.Model):
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objects = BaseManager.from_queryset(CustomQuerySet)()
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You may also store the generated class into a variable::
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CustomManager = BaseManager.from_queryset(CustomQuerySet)
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class MyModel(models.Model):
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objects = CustomManager()
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.. _custom-managers-and-inheritance:
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Custom managers and model inheritance
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-------------------------------------
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Here's how Django handles custom managers and :ref:`model inheritance
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<model-inheritance>`:
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#. Managers from base classes are always inherited by the child class,
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using Python's normal name resolution order (names on the child
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class override all others; then come names on the first parent class,
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and so on).
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#. If no managers are declared on a model and/or its parents, Django
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automatically creates the ``objects`` manager.
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#. The default manager on a class is either the one chosen with
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:attr:`Meta.default_manager_name
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<django.db.models.Options.default_manager_name>`, or the first manager
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declared on the model, or the default manager of the first parent model.
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These rules provide the necessary flexibility if you want to install a
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collection of custom managers on a group of models, via an abstract base
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class, but still customize the default manager. For example, suppose you have
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this base class::
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class AbstractBase(models.Model):
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# ...
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objects = CustomManager()
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class Meta:
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abstract = True
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If you use this directly in a subclass, ``objects`` will be the default
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manager if you declare no managers in the base class::
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class ChildA(AbstractBase):
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# ...
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# This class has CustomManager as the default manager.
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pass
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If you want to inherit from ``AbstractBase``, but provide a different default
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manager, you can provide the default manager on the child class::
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class ChildB(AbstractBase):
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# ...
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# An explicit default manager.
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default_manager = OtherManager()
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Here, ``default_manager`` is the default. The ``objects`` manager is
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still available, since it's inherited, but isn't used as the default.
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Finally for this example, suppose you want to add extra managers to the child
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class, but still use the default from ``AbstractBase``. You can't add the new
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manager directly in the child class, as that would override the default and you would
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have to also explicitly include all the managers from the abstract base class.
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The solution is to put the extra managers in another base class and introduce
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it into the inheritance hierarchy *after* the defaults::
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class ExtraManager(models.Model):
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extra_manager = OtherManager()
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class Meta:
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abstract = True
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class ChildC(AbstractBase, ExtraManager):
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# ...
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# Default manager is CustomManager, but OtherManager is
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# also available via the "extra_manager" attribute.
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pass
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Note that while you can *define* a custom manager on the abstract model, you
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can't *invoke* any methods using the abstract model. That is::
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ClassA.objects.do_something()
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is legal, but::
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AbstractBase.objects.do_something()
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will raise an exception. This is because managers are intended to encapsulate
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logic for managing collections of objects. Since you can't have a collection of
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abstract objects, it doesn't make sense to be managing them. If you have
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functionality that applies to the abstract model, you should put that functionality
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in a ``staticmethod`` or ``classmethod`` on the abstract model.
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Implementation concerns
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-----------------------
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Whatever features you add to your custom ``Manager``, it must be
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possible to make a shallow copy of a ``Manager`` instance; i.e., the
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following code must work::
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>>> import copy
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>>> manager = MyManager()
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>>> my_copy = copy.copy(manager)
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Django makes shallow copies of manager objects during certain queries;
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if your Manager cannot be copied, those queries will fail.
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This won't be an issue for most custom managers. If you are just
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adding simple methods to your ``Manager``, it is unlikely that you
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will inadvertently make instances of your ``Manager`` uncopyable.
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However, if you're overriding ``__getattr__`` or some other private
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method of your ``Manager`` object that controls object state, you
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should ensure that you don't affect the ability of your ``Manager`` to
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be copied.
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