django/docs/sites.txt

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

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 whole sites framework is based on two simple concepts:

    * The ``Site`` model, found in ``django.contrib.sites``, has ``domain`` and
      ``name`` fields.
    * The ``SITE_ID`` setting specifies the database ID of the ``Site`` object
      associated with that particular settings file.

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 ``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.conf import settings

          def article_detail(request, article_id):
              try:
                  a = Article.objects.get(id=article_id, sites__id__exact=settings.SITE_ID)
              except Article.DoesNotExist:
                  raise Http404
              # ...

.. _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 ``Site`` model in a many-to-one
relationship, using ``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 the current site from views
----------------------------------------

On a lower level, you can use the sites framework in your Django views to do
particular things based on what 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.
        else:
            # Do something else.

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. A slightly
cleaner way of accomplishing the same thing is to check the current site's
domain::

    from django.conf import settings
    from django.contrib.sites.models import Site

    def my_view(request):
        current_site = Site.objects.get(id=settings.SITE_ID)
        if current_site.domain == 'foo.com':
            # Do something
        else:
            # Do something else.

The idiom of retrieving the ``Site`` object for the value of
``settings.SITE_ID`` is quite common, so the ``Site`` model's manager has a
``get_current()`` method. This example is equivalent to the previous one::

    from django.contrib.sites.models import Site

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

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

LJWorld.com and Lawrence.com both have e-mail 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 he immediately gets an e-mail saying,
"Thanks for your subscription."

It'd be inefficient and redundant to implement this signup-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 ``Site``
objects, we can abstract the "thank you" notice to use the values of the
current site's ``name`` and ``domain``.

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

    from django.contrib.sites.models import Site
    from django.core.mail import send_mail

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

        current_site = Site.objects.get_current()
        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 e-mail has the subject line "Thanks for subscribing to
lawrence.com alerts." On LJWorld.com, the e-mail has the subject "Thanks for
subscribing to LJWorld.com alerts." Same goes for the e-mail'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 (``TEMPLATE_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 ``subject.txt`` and ``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 ``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/'

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

**New in Django development version**

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``
==========================

If ``Site``\s play a key role in your application, consider using the helpful
``CurrentSiteManager`` in your model(s). It's a model manager_ that
automatically filters its queries to include only objects associated with the
current ``Site``.

Use ``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 ``SITE_ID`` setting.

Put another way, these two statements are equivalent::

    Photo.objects.filter(site=settings.SITE_ID)
    Photo.on_site.all()

How did ``CurrentSiteManager`` know which field of ``Photo`` was the ``Site``?
It defaults to looking for a field called ``site``. If your model has a
``ForeignKey`` or ``ManyToManyField`` called something *other* than ``site``,
you need to explicitly pass that as the parameter to ``CurrentSiteManager``.
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 ``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 ``CurrentSiteManager``. As explained
in the `manager documentation`_, 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
``CurrentSiteManager``.

.. _manager: ../model-api/#managers
.. _manager documentation: ../model-api/#managers

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 ``SITE_ID`` setting.

Here's how Django uses the sites framework:

    * In the `redirects framework`_, each redirect object is associated with a
      particular site. When Django searches for a redirect, it takes into
      account the current ``SITE_ID``.

    * In the comments framework, each comment is associated with a particular
      site. When a comment is posted, its ``site`` is set to the current
      ``SITE_ID``, and when comments are listed via the appropriate template
      tag, only the comments for the current site are displayed.

    * In the `flatpages framework`_, each flatpage is associated with a
      particular site. When a flatpage is created, you specify its ``site``,
      and the ``FlatpageFallbackMiddleware`` checks the current ``SITE_ID`` in
      retrieving flatpages to display.

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

    * In the `authentication framework`_, the ``django.contrib.auth.views.login``
      view passes the current ``Site`` name to the template as ``{{{ site_name }}}``.

    * The shortcut view (``django.views.defaults.shortcut``) uses the domain of
      the current ``Site`` object when calculating an object's URL.

    * In the admin framework, the ''view on site'' link uses the current
      ``Site`` to work out the domain for the site that it will redirect to.

.. _redirects framework: ../redirects/
.. _flatpages framework: ../flatpages/
.. _syndication framework: ../syndication_feeds/
.. _authentication framework: ../authentication/

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

**New in Django development version**

Some ``django.contrib`` 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 ``RequestSite`` class, which can be used
as a fallback when the database-backed sites framework is not available.

A ``RequestSite`` object has a similar interface to a normal ``Site`` object,
except its ``__init__()`` method takes an ``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 ``Site``, but
the methods raise ``NotImplementedError``.