[1.1.X] Fixed #12040 -- Cleaned up the documentation describing how CurrentSiteManager works. Thanks to elpaso66 for the report, and Gabriel Hurley for the draft patch.
Backport of r13168 from trunk. git-svn-id: http://code.djangoproject.com/svn/django/branches/releases/1.1.X@13174 bcc190cf-cafb-0310-a4f2-bffc1f526a37
This commit is contained in:
parent
3b9a82714c
commit
814268fb54
|
@ -257,12 +257,12 @@ The ``CurrentSiteManager``
|
|||
|
||||
.. class:: django.contrib.sites.managers.CurrentSiteManager
|
||||
|
||||
If :class:`~django.contrib.sites.models.Site`\s play a key role in your application,
|
||||
consider using the helpful
|
||||
:class:`~django.contrib.sites.managers.CurrentSiteManager` in your model(s).
|
||||
It's a model :ref:`manager <topics-db-managers>` that automatically filters
|
||||
its queries to include only objects associated with the current
|
||||
:class:`~django.contrib.sites.models.Site`.
|
||||
If :class:`~django.contrib.sites.models.Site` plays a key role in your
|
||||
application, consider using the helpful
|
||||
:class:`~django.contrib.sites.managers.CurrentSiteManager` in your
|
||||
model(s). It's a model :ref:`manager <topics-db-managers>` that
|
||||
automatically filters its queries to include only objects associated
|
||||
with the current :class:`~django.contrib.sites.models.Site`.
|
||||
|
||||
Use :class:`~django.contrib.sites.managers.CurrentSiteManager` by adding it to
|
||||
your model explicitly. For example::
|
||||
|
@ -288,16 +288,21 @@ Put another way, these two statements are equivalent::
|
|||
Photo.objects.filter(site=settings.SITE_ID)
|
||||
Photo.on_site.all()
|
||||
|
||||
How did :class:`~django.contrib.sites.managers.CurrentSiteManager` know which
|
||||
field of ``Photo`` was the :class:`~django.contrib.sites.models.Site`? It
|
||||
defaults to looking for a field called
|
||||
:class:`~django.contrib.sites.models.Site`. If your model has a
|
||||
:class:`~django.db.models.fields.related.ForeignKey` or
|
||||
:class:`~django.db.models.fields.related.ManyToManyField` called something
|
||||
*other* than :class:`~django.contrib.sites.models.Site`, you need to explicitly
|
||||
pass that as the parameter to
|
||||
:class:`~django.contrib.sites.managers.CurrentSiteManager`. The following model,
|
||||
which has a field called ``publish_on``, demonstrates this::
|
||||
How did :class:`~django.contrib.sites.managers.CurrentSiteManager`
|
||||
know which field of ``Photo`` was the
|
||||
:class:`~django.contrib.sites.models.Site`? By default,
|
||||
:class:`~django.contrib.sites.managers.CurrentSiteManager` looks for a
|
||||
either a :class:`~django.db.models.fields.related.ForeignKey` called
|
||||
``site`` or a
|
||||
:class:`~django.db.models.fields.related.ManyToManyField` called
|
||||
``sites`` to filter on. If you use a field named something other than
|
||||
``site`` or ``sites`` to identify which
|
||||
:class:`~django.contrib.sites.models.Site` objects your object is
|
||||
related to, then you need to explicitly pass the custom field name as
|
||||
a parameter to
|
||||
:class:`~django.contrib.sites.managers.CurrentSiteManager` on your
|
||||
model. The following model, which has a field called ``publish_on``,
|
||||
demonstrates this::
|
||||
|
||||
from django.db import models
|
||||
from django.contrib.sites.models import Site
|
||||
|
@ -314,16 +319,17 @@ which has a field called ``publish_on``, demonstrates this::
|
|||
If you attempt to use :class:`~django.contrib.sites.managers.CurrentSiteManager`
|
||||
and pass a field name that doesn't exist, Django will raise a :exc:`ValueError`.
|
||||
|
||||
Finally, note that you'll probably want to keep a normal (non-site-specific)
|
||||
``Manager`` on your model, even if you use
|
||||
:class:`~django.contrib.sites.managers.CurrentSiteManager`. As explained
|
||||
in the :ref:`manager documentation <topics-db-managers>`, if you define a manager
|
||||
manually, then Django won't create the automatic ``objects = models.Manager()``
|
||||
manager for you.Also, note that certain parts of Django -- namely, the Django admin site and
|
||||
generic views -- use whichever manager is defined *first* in the model, so if
|
||||
you want your admin site to have access to all objects (not just site-specific
|
||||
ones), put ``objects = models.Manager()`` in your model, before you define
|
||||
:class:`~django.contrib.sites.managers.CurrentSiteManager`.
|
||||
Finally, note that you'll probably want to keep a normal
|
||||
(non-site-specific) ``Manager`` on your model, even if you use
|
||||
:class:`~django.contrib.sites.managers.CurrentSiteManager`. As
|
||||
explained in the :ref:`manager documentation <topics-db-managers>`, if
|
||||
you define a manager manually, then Django won't create the automatic
|
||||
``objects = models.Manager()`` manager for you. Also note that certain
|
||||
parts of Django -- namely, the Django admin site and generic views --
|
||||
use whichever manager is defined *first* in the model, so if you want
|
||||
your admin site to have access to all objects (not just site-specific
|
||||
ones), put ``objects = models.Manager()`` in your model, before you
|
||||
define :class:`~django.contrib.sites.managers.CurrentSiteManager`.
|
||||
|
||||
How Django uses the sites framework
|
||||
===================================
|
||||
|
@ -345,7 +351,7 @@ Here's how Django uses the sites framework:
|
|||
:class:`~django.contrib.sites.models.Site` is set to the current
|
||||
:setting:`SITE_ID`, and when comments are listed via the appropriate
|
||||
template tag, only the comments for the current site are displayed.
|
||||
|
||||
|
||||
* In the :mod:`flatpages framework <django.contrib.flatpages>`, each
|
||||
flatpage is associated with a particular site. When a flatpage is created,
|
||||
you specify its :class:`~django.contrib.sites.models.Site`, and the
|
||||
|
|
Loading…
Reference in New Issue