Added documentation for CurrentSiteManager to docs/sites.txt

    >>> 'http://%s%s' % (Site.objects.get_current().domain, obj.get_absolute_url())

    'http://example.com/mymodel/objects/3/'

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(maxlength=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.

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(maxlength=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``.

.. _manager: http://www.djangoproject.com/documentation/model_api/#managers

How Django uses the sites framework

===================================