Added documentation for CurrentSiteManager to docs/sites.txt
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@ -213,6 +213,56 @@ To do this, you can use the sites framework. A simple example::
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>>> 'http://%s%s' % (Site.objects.get_current().domain, obj.get_absolute_url())
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'http://example.com/mymodel/objects/3/'
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The ``CurrentSiteManager``
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==========================
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If ``Site``s play a key role in your application, consider using the helpful
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``CurrentSiteManager`` in your model(s). It's a model manager_ that
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automatically filters its queries to include only objects associated with the
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current ``Site``.
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Use ``CurrentSiteManager`` by adding it to your model explicitly. For example::
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from django.db import models
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from django.contrib.sites.models import Site
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from django.contrib.sites.managers import CurrentSiteManager
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class Photo(models.Model):
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photo = models.FileField(upload_to='/home/photos')
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photographer_name = models.CharField(maxlength=100)
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pub_date = models.DateField()
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site = models.ForeignKey(Site)
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objects = models.Manager()
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on_site = CurrentSiteManager()
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With this model, ``Photo.objects.all()`` will return all ``Photo`` objects in
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the database, but ``Photo.on_site.all()`` will return only the ``Photo``
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objects associated with the current site, according to the ``SITE_ID`` setting.
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How did ``CurrentSiteManager`` know which field of ``Photo`` was the ``Site``?
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It defaults to looking for a field called ``site``. If your model has a
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``ForeignKey`` or ``ManyToManyField`` called something *other* than ``site``,
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you need to explicitly pass that as the parameter to ``CurrentSiteManager``.
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The following model, which has a field called ``publish_on``, demonstrates
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this::
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from django.db import models
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from django.contrib.sites.models import Site
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from django.contrib.sites.managers import CurrentSiteManager
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class Photo(models.Model):
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photo = models.FileField(upload_to='/home/photos')
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photographer_name = models.CharField(maxlength=100)
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pub_date = models.DateField()
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publish_on = models.ForeignKey(Site)
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objects = models.Manager()
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on_site = CurrentSiteManager('publish_on')
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If you attempt to use ``CurrentSiteManager`` and pass a field name that doesn't
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exist, Django will raise a ``ValueError``.
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.. _manager: http://www.djangoproject.com/documentation/model_api/#managers
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How Django uses the sites framework
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===================================
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