django/docs/ref/contrib/sites.txt

=====================
The "sites" framework
=====================

.. module:: django.contrib.sites
   :synopsis: Lets you operate multiple Web sites from the same database and
              Django project

Django comes with an optional "sites" framework. It's a hook for associating
objects and functionality to particular Web sites, and it's a holding place for
the domain names and "verbose" names of your Django-powered sites.

Use it if your single Django installation powers more than one site and you
need to differentiate between those sites in some way.

The sites framework is mainly based on a simple model:

.. class:: models.Site

    A model for storing the ``domain`` and ``name`` attributes of a Web site.

    .. attribute:: domain

        The domain name associated with the Web site.

    .. attribute:: name

        A human-readable "verbose" name for the Web site.

The :setting:`SITE_ID` setting specifies the database ID of the
:class:`~django.contrib.sites.models.Site` object associated with that
particular settings file. If the setting is omitted, the
:func:`~django.contrib.sites.shortcuts.get_current_site` function will
try to get the current site by comparing the
:attr:`~django.contrib.sites.models.Site.domain` with the host name from
the :meth:`request.get_host() <django.http.HttpRequest.get_host>` method.

How you use this is up to you, but Django uses it in a couple of ways
automatically via simple conventions.

Example usage
=============

Why would you use sites? It's best explained through examples.

Associating content with multiple sites
---------------------------------------

The Django-powered sites LJWorld.com_ and Lawrence.com_ are operated by the
same news organization -- the Lawrence Journal-World newspaper in Lawrence,
Kansas. LJWorld.com focuses on news, while Lawrence.com focuses on local
entertainment. But sometimes editors want to publish an article on *both*
sites.

The brain-dead way of solving the problem would be to require site producers to
publish the same story twice: once for LJWorld.com and again for Lawrence.com.
But that's inefficient for site producers, and it's redundant to store
multiple copies of the same story in the database.

The better solution is simple: Both sites use the same article database, and an
article is associated with one or more sites. In Django model terminology,
that's represented by a :class:`~django.db.models.ManyToManyField` in the
``Article`` model::

    from django.db import models
    from django.contrib.sites.models import Site

    class Article(models.Model):
        headline = models.CharField(max_length=200)
        # ...
        sites = models.ManyToManyField(Site)

This accomplishes several things quite nicely:

* It lets the site producers edit all content -- on both sites -- in a
  single interface (the Django admin).

* It means the same story doesn't have to be published twice in the
  database; it only has a single record in the database.

* It lets the site developers use the same Django view code for both sites.
  The view code that displays a given story just checks to make sure the
  requested story is on the current site. It looks something like this::

      from django.contrib.sites.shortcuts import get_current_site

      def article_detail(request, article_id):
          try:
              a = Article.objects.get(id=article_id, sites__id=get_current_site(request).id)
          except Article.DoesNotExist:
              raise Http404("Article does not exist on this site")
          # ...

.. _ljworld.com: http://www.ljworld.com/
.. _lawrence.com: http://www.lawrence.com/

Associating content with a single site
--------------------------------------

Similarly, you can associate a model to the
:class:`~django.contrib.sites.models.Site`
model in a many-to-one relationship, using
:class:`~django.db.models.ForeignKey`.

For example, if an article is only allowed on a single site, you'd use a model
like this::

    from django.db import models
    from django.contrib.sites.models import Site

    class Article(models.Model):
        headline = models.CharField(max_length=200)
        # ...
        site = models.ForeignKey(Site)

This has the same benefits as described in the last section.

.. _hooking-into-current-site-from-views:

Hooking into the current site from views
----------------------------------------

You can use the sites framework in your Django views to do
particular things based on the site in which the view is being called.
For example::

    from django.conf import settings

    def my_view(request):
        if settings.SITE_ID == 3:
            # Do something.
            pass
        else:
            # Do something else.
            pass

Of course, it's ugly to hard-code the site IDs like that. This sort of
hard-coding is best for hackish fixes that you need done quickly. The
cleaner way of accomplishing the same thing is to check the current site's
domain::

    from django.contrib.sites.shortcuts import get_current_site

    def my_view(request):
        current_site = get_current_site(request)
        if current_site.domain == 'foo.com':
            # Do something
            pass
        else:
            # Do something else.
            pass

This has also the advantage of checking if the sites framework is installed,
and return a :class:`~django.contrib.sites.requests.RequestSite` instance if
it is not.

If you don't have access to the request object, you can use the
``get_current()`` method of the :class:`~django.contrib.sites.models.Site`
model's manager. You should then ensure that your settings file does contain
the :setting:`SITE_ID` setting. This example is equivalent to the previous one::

    from django.contrib.sites.models import Site

    def my_function_without_request():
        current_site = Site.objects.get_current()
        if current_site.domain == 'foo.com':
            # Do something
            pass
        else:
            # Do something else.
            pass

Getting the current domain for display
--------------------------------------

LJWorld.com and Lawrence.com both have email alert functionality, which lets
readers sign up to get notifications when news happens. It's pretty basic: A
reader signs up on a Web form and immediately gets an email saying,
"Thanks for your subscription."

It'd be inefficient and redundant to implement this sign up processing code
twice, so the sites use the same code behind the scenes. But the "thank you for
signing up" notice needs to be different for each site. By using
:class:`~django.contrib.sites.models.Site`
objects, we can abstract the "thank you" notice to use the values of the
current site's :attr:`~django.contrib.sites.models.Site.name` and
:attr:`~django.contrib.sites.models.Site.domain`.

Here's an example of what the form-handling view looks like::

    from django.contrib.sites.shortcuts import get_current_site
    from django.core.mail import send_mail

    def register_for_newsletter(request):
        # Check form values, etc., and subscribe the user.
        # ...

        current_site = get_current_site(request)
        send_mail('Thanks for subscribing to %s alerts' % current_site.name,
            'Thanks for your subscription. We appreciate it.\n\n-The %s team.' % current_site.name,
            'editor@%s' % current_site.domain,
            [user.email])

        # ...

On Lawrence.com, this email has the subject line "Thanks for subscribing to
lawrence.com alerts." On LJWorld.com, the email has the subject "Thanks for
subscribing to LJWorld.com alerts." Same goes for the email's message body.

Note that an even more flexible (but more heavyweight) way of doing this would
be to use Django's template system. Assuming Lawrence.com and LJWorld.com have
different template directories (:setting:`DIRS <TEMPLATES-DIRS>`), you could
simply farm out to the template system like so::

    from django.core.mail import send_mail
    from django.template import loader, Context

    def register_for_newsletter(request):
        # Check form values, etc., and subscribe the user.
        # ...

        subject = loader.get_template('alerts/subject.txt').render(Context({}))
        message = loader.get_template('alerts/message.txt').render(Context({}))
        send_mail(subject, message, 'editor@ljworld.com', [user.email])

        # ...

In this case, you'd have to create :file:`subject.txt` and :file:`message.txt`
template files for both the LJWorld.com and Lawrence.com template directories.
That gives you more flexibility, but it's also more complex.

It's a good idea to exploit the :class:`~django.contrib.sites.models.Site`
objects as much as possible, to remove unneeded complexity and redundancy.

Getting the current domain for full URLs
----------------------------------------

Django's ``get_absolute_url()`` convention is nice for getting your objects'
URL without the domain name, but in some cases you might want to display the
full URL -- with ``http://`` and the domain and everything -- for an object.
To do this, you can use the sites framework. A simple example::

    >>> from django.contrib.sites.models import Site
    >>> obj = MyModel.objects.get(id=3)
    >>> obj.get_absolute_url()
    '/mymodel/objects/3/'
    >>> Site.objects.get_current().domain
    'example.com'
    >>> 'http://%s%s' % (Site.objects.get_current().domain, obj.get_absolute_url())
    'http://example.com/mymodel/objects/3/'

.. _enabling-the-sites-framework:

Enabling the sites framework
============================

To enable the sites framework, follow these steps:

1. Add ``'django.contrib.sites'`` to your :setting:`INSTALLED_APPS`
   setting.

2. Define a :setting:`SITE_ID` setting::

    SITE_ID = 1

3. Run :djadmin:`migrate`.

``django.contrib.sites`` registers a
:data:`~django.db.models.signals.post_migrate` signal handler which creates a
default site named ``example.com`` with the domain ``example.com``. This site
will also be created after Django creates the test database. To set the
correct name and domain for your project, you can use a :ref:`data migration
<data-migrations>`.

In order to serve different sites in production, you'd create a separate
settings file with each ``SITE_ID`` (perhaps importing from a common settings
file to avoid duplicating shared settings) and then specify the appropriate
:envvar:`DJANGO_SETTINGS_MODULE` for each site.

Caching the current ``Site`` object
===================================

As the current site is stored in the database, each call to
``Site.objects.get_current()`` could result in a database query. But Django is a
little cleverer than that: on the first request, the current site is cached, and
any subsequent call returns the cached data instead of hitting the database.

If for any reason you want to force a database query, you can tell Django to
clear the cache using ``Site.objects.clear_cache()``::

    # First call; current site fetched from database.
    current_site = Site.objects.get_current()
    # ...

    # Second call; current site fetched from cache.
    current_site = Site.objects.get_current()
    # ...

    # Force a database query for the third call.
    Site.objects.clear_cache()
    current_site = Site.objects.get_current()

The ``CurrentSiteManager``
==========================

.. class:: managers.CurrentSiteManager

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 :doc:`manager </topics/db/managers>` that
automatically filters its queries to include only objects associated
with the current :class:`~django.contrib.sites.models.Site`.

.. admonition:: Mandatory :setting:`SITE_ID`

    The ``CurrentSiteManager`` is only usable when the :setting:`SITE_ID`
    setting is defined in your settings.

Use :class:`~django.contrib.sites.managers.CurrentSiteManager` by adding it to
your model explicitly. For example::

    from django.db import models
    from django.contrib.sites.models import Site
    from django.contrib.sites.managers import CurrentSiteManager

    class Photo(models.Model):
        photo = models.FileField(upload_to='/home/photos')
        photographer_name = models.CharField(max_length=100)
        pub_date = models.DateField()
        site = models.ForeignKey(Site)
        objects = models.Manager()
        on_site = CurrentSiteManager()

With this model, ``Photo.objects.all()`` will return all ``Photo`` objects in
the database, but ``Photo.on_site.all()`` will return only the ``Photo`` objects
associated with the current site, according to the :setting:`SITE_ID` setting.

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`? By default,
:class:`~django.contrib.sites.managers.CurrentSiteManager` looks for a
either a :class:`~django.db.models.ForeignKey` called
``site`` or a
:class:`~django.db.models.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
    from django.contrib.sites.managers import CurrentSiteManager

    class Photo(models.Model):
        photo = models.FileField(upload_to='/home/photos')
        photographer_name = models.CharField(max_length=100)
        pub_date = models.DateField()
        publish_on = models.ForeignKey(Site)
        objects = models.Manager()
        on_site = CurrentSiteManager('publish_on')

If you attempt to use :class:`~django.contrib.sites.managers.CurrentSiteManager`
and pass a field name that doesn't exist, Django will raise a ``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 :doc:`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`.

.. _site-middleware:

Site middleware
===============

.. versionadded:: 1.7

If you often use this pattern::

    from django.contrib.sites.models import Site

    def my_view(request):
        site = Site.objects.get_current()
        ...

there is simple way to avoid repetitions. Add
:class:`django.contrib.sites.middleware.CurrentSiteMiddleware` to
:setting:`MIDDLEWARE_CLASSES`. The middleware sets the ``site`` attribute on
every request object, so you can use ``request.site`` to get the current site.

How Django uses the sites framework
===================================

Although it's not required that you use the sites framework, it's strongly
encouraged, because Django takes advantage of it in a few places. Even if your
Django installation is powering only a single site, you should take the two
seconds to create the site object with your ``domain`` and ``name``, and point
to its ID in your :setting:`SITE_ID` setting.

Here's how Django uses the sites framework:

* In the :mod:`redirects framework <django.contrib.redirects>`, each
  redirect object is associated with a particular site. When Django searches
  for a redirect, it takes into account the current site.

* 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
  :class:`~django.contrib.flatpages.middleware.FlatpageFallbackMiddleware`
  checks the current site in retrieving flatpages to display.

* In the :mod:`syndication framework <django.contrib.syndication>`, the
  templates for ``title`` and ``description`` automatically have access to a
  variable ``{{ site }}``, which is the
  :class:`~django.contrib.sites.models.Site` object representing the current
  site. Also, the hook for providing item URLs will use the ``domain`` from
  the current :class:`~django.contrib.sites.models.Site` object if you don't
  specify a fully-qualified domain.

* In the :mod:`authentication framework <django.contrib.auth>`, the
  :func:`django.contrib.auth.views.login` view passes the current
  :class:`~django.contrib.sites.models.Site` name to the template as
  ``{{ site_name }}``.

* The shortcut view (``django.contrib.contenttypes.views.shortcut``)
  uses the domain of the current
  :class:`~django.contrib.sites.models.Site` object when calculating
  an object's URL.

* In the admin framework, the "view on site" link uses the current
  :class:`~django.contrib.sites.models.Site` to work out the domain for the
  site that it will redirect to.

``RequestSite`` objects
=======================

.. _requestsite-objects:

Some :doc:`django.contrib </ref/contrib/index>` applications take advantage of
the sites framework but are architected in a way that doesn't *require* the
sites framework to be installed in your database. (Some people don't want to,
or just aren't *able* to install the extra database table that the sites
framework requires.) For those cases, the framework provides a
:class:`django.contrib.sites.requests.RequestSite` class, which can be used as
a fallback when the database-backed sites framework is not available.

.. class:: requests.RequestSite

    A class that shares the primary interface of
    :class:`~django.contrib.sites.models.Site` (i.e., it has
    ``domain`` and ``name`` attributes) but gets its data from a Django
    :class:`~django.http.HttpRequest` object rather than from a database.

    .. method:: __init__(request)

        Sets the ``name`` and ``domain`` attributes to the value of
        :meth:`~django.http.HttpRequest.get_host`.

    .. deprecated:: 1.7

        This class used to be defined in ``django.contrib.sites.models``. The
        old import location will work until Django 1.9.

A :class:`~django.contrib.sites.requests.RequestSite` object has a similar
interface to a normal :class:`~django.contrib.sites.models.Site` object,
except its :meth:`~django.contrib.sites.requests.RequestSite.__init__()`
method takes an :class:`~django.http.HttpRequest` object. It's able to deduce
the ``domain`` and ``name`` by looking at the request's domain. It has
``save()`` and ``delete()`` methods to match the interface of
:class:`~django.contrib.sites.models.Site`, but the methods raise
:exc:`NotImplementedError`.

``get_current_site`` shortcut
=============================

Finally, to avoid repetitive fallback code, the framework provides a
:func:`django.contrib.sites.shortcuts.get_current_site` function.

.. function:: shortcuts.get_current_site(request)

    A function that checks if ``django.contrib.sites`` is installed and
    returns either the current :class:`~django.contrib.sites.models.Site`
    object or a :class:`~django.contrib.sites.requests.RequestSite` object
    based on the request.

    .. deprecated:: 1.7

        This function used to be defined in ``django.contrib.sites.models``.
        The old import location will work until Django 1.9.

    .. versionchanged:: 1.8

        This function will now lookup the current site based on
        :meth:`request.get_host() <django.http.HttpRequest.get_host>` if the
        :setting:`SITE_ID` setting is not defined.