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312 lines
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312 lines
12 KiB
Plaintext
=====================
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The "sites" framework
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=====================
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Django comes with an optional "sites" framework. It's a hook for associating
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objects and functionality to particular Web sites, and it's a holding place for
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the domain names and "verbose" names of your Django-powered sites.
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Use it if your single Django installation powers more than one site and you
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need to differentiate between those sites in some way.
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The whole sites framework is based on two simple concepts:
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* The ``Site`` model, found in ``django.contrib.sites``, has ``domain`` and
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``name`` fields.
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* The ``SITE_ID`` setting specifies the database ID of the ``Site`` object
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associated with that particular settings file.
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How you use this is up to you, but Django uses it in a couple of ways
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automatically via simple conventions.
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Example usage
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=============
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Why would you use sites? It's best explained through examples.
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Associating content with multiple sites
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---------------------------------------
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The Django-powered sites LJWorld.com_ and Lawrence.com_ are operated by the
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same news organization -- the Lawrence Journal-World newspaper in Lawrence,
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Kansas. LJWorld.com focuses on news, while Lawrence.com focuses on local
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entertainment. But sometimes editors want to publish an article on *both*
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sites.
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The brain-dead way of solving the problem would be to require site producers to
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publish the same story twice: once for LJWorld.com and again for Lawrence.com.
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But that's inefficient for site producers, and it's redundant to store
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multiple copies of the same story in the database.
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The better solution is simple: Both sites use the same article database, and an
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article is associated with one or more sites. In Django model terminology,
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that's represented by a ``ManyToManyField`` in the ``Article`` model::
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from django.db import models
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from django.contrib.sites.models import Site
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class Article(models.Model):
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headline = models.CharField(maxlength=200)
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# ...
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sites = models.ManyToManyField(Site)
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This accomplishes several things quite nicely:
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* It lets the site producers edit all content -- on both sites -- in a
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single interface (the Django admin).
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* It means the same story doesn't have to be published twice in the
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database; it only has a single record in the database.
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* It lets the site developers use the same Django view code for both sites.
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The view code that displays a given story just checks to make sure the
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requested story is on the current site. It looks something like this::
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from django.conf import settings
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def article_detail(request, article_id):
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try:
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a = Article.objects.get(id=article_id, sites__id__exact=settings.SITE_ID)
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except Article.DoesNotExist:
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raise Http404
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# ...
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.. _ljworld.com: http://www.ljworld.com/
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.. _lawrence.com: http://www.lawrence.com/
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Associating content with a single site
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--------------------------------------
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Similarly, you can associate a model to the ``Site`` model in a many-to-one
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relationship, using ``ForeignKey``.
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For example, if an article is only allowed on a single site, you'd use a model
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like 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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class Article(models.Model):
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headline = models.CharField(maxlength=200)
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# ...
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site = models.ForeignKey(Site)
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This has the same benefits as described in the last section.
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Hooking into the current site from views
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----------------------------------------
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On a lower level, you can use the sites framework in your Django views to do
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particular things based on what site in which the view is being called.
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For example::
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from django.conf import settings
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def my_view(request):
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if settings.SITE_ID == 3:
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# Do something.
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else:
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# Do something else.
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Of course, it's ugly to hard-code the site IDs like that. This sort of
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hard-coding is best for hackish fixes that you need done quickly. A slightly
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cleaner way of accomplishing the same thing is to check the current site's
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domain::
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from django.conf import settings
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from django.contrib.sites.models import Site
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def my_view(request):
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current_site = Site.objects.get(id=settings.SITE_ID)
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if current_site.domain == 'foo.com':
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# Do something
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else:
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# Do something else.
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The idiom of retrieving the ``Site`` object for the value of
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``settings.SITE_ID`` is quite common, so the ``Site`` model's manager has a
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``get_current()`` method. This example is equivalent to the previous one::
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from django.contrib.sites.models import Site
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def my_view(request):
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current_site = Site.objects.get_current()
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if current_site.domain == 'foo.com':
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# Do something
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else:
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# Do something else.
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Getting the current domain for display
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--------------------------------------
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LJWorld.com and Lawrence.com both have e-mail alert functionality, which lets
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readers sign up to get notifications when news happens. It's pretty basic: A
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reader signs up on a Web form, and he immediately gets an e-mail saying,
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"Thanks for your subscription."
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It'd be inefficient and redundant to implement this signup-processing code
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twice, so the sites use the same code behind the scenes. But the "thank you for
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signing up" notice needs to be different for each site. By using ``Site``
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objects, we can abstract the "thank you" notice to use the values of the
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current site's ``name`` and ``domain``.
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Here's an example of what the form-handling view looks like::
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from django.contrib.sites.models import Site
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from django.core.mail import send_mail
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def register_for_newsletter(request):
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# Check form values, etc., and subscribe the user.
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# ...
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current_site = Site.objects.get_current()
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send_mail('Thanks for subscribing to %s alerts' % current_site.name,
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'Thanks for your subscription. We appreciate it.\n\n-The %s team.' % current_site.name,
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'editor@%s' % current_site.domain,
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[user.email])
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# ...
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On Lawrence.com, this e-mail has the subject line "Thanks for subscribing to
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lawrence.com alerts." On LJWorld.com, the e-mail has the subject "Thanks for
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subscribing to LJWorld.com alerts." Same goes for the e-mail's message body.
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Note that an even more flexible (but more heavyweight) way of doing this would
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be to use Django's template system. Assuming Lawrence.com and LJWorld.com have
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different template directories (``TEMPLATE_DIRS``), you could simply farm out
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to the template system like so::
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from django.core.mail import send_mail
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from django.template import loader, Context
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def register_for_newsletter(request):
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# Check form values, etc., and subscribe the user.
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# ...
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subject = loader.get_template('alerts/subject.txt').render(Context({}))
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message = loader.get_template('alerts/message.txt').render(Context({}))
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send_mail(subject, message, 'editor@ljworld.com', [user.email])
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# ...
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In this case, you'd have to create ``subject.txt`` and ``message.txt`` template
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files for both the LJWorld.com and Lawrence.com template directories. That
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gives you more flexibility, but it's also more complex.
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It's a good idea to exploit the ``Site`` objects as much as possible, to remove
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unneeded complexity and redundancy.
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Getting the current domain for full URLs
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----------------------------------------
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Django's ``get_absolute_url()`` convention is nice for getting your objects'
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URL without the domain name, but in some cases you might want to display the
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full URL -- with ``http://`` and the domain and everything -- for an object.
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To do this, you can use the sites framework. A simple example::
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>>> from django.contrib.sites.models import Site
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>>> obj = MyModel.objects.get(id=3)
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>>> obj.get_absolute_url()
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'/mymodel/objects/3/'
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>>> Site.objects.get_current().domain
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'example.com'
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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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Put another way, these two statements are equivalent::
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Photo.objects.filter(site=settings.SITE_ID)
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Photo.on_site.all()
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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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Although it's not required that you use the sites framework, it's strongly
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encouraged, because Django takes advantage of it in a few places. Even if your
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Django installation is powering only a single site, you should take the two
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seconds to create the site object with your ``domain`` and ``name``, and point
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to its ID in your ``SITE_ID`` setting.
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Here's how Django uses the sites framework:
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* In the `redirects framework`_, each redirect object is associated with a
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particular site. When Django searches for a redirect, it takes into
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account the current ``SITE_ID``.
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* In the comments framework, each comment is associated with a particular
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site. When a comment is posted, its ``site`` is set to the current
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``SITE_ID``, and when comments are listed via the appropriate template
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tag, only the comments for the current site are displayed.
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* In the `flatpages framework`_, each flatpage is associated with a
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particular site. When a flatpage is created, you specify its ``site``,
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and the ``FlatpageFallbackMiddleware`` checks the current ``SITE_ID`` in
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retrieving flatpages to display.
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* In the `syndication framework`_, the templates for ``title`` and
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``description`` automatically have access to a variable ``{{{ site }}}``,
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which is the ``Site`` object representing the current site. Also, the
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hook for providing item URLs will use the ``domain`` from the current
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``Site`` object if you don't specify a fully-qualified domain.
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* In the `authentication framework`_, the ``django.contrib.auth.views.login``
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view passes the current ``Site`` name to the template as ``{{{ site_name }}}``.
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* The shortcut view (``django.views.defaults.shortcut``) uses the domain of
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the current ``Site`` object when calculating an object's URL.
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.. _redirects framework: http://www.djangoproject.com/documentation/redirects/
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.. _flatpages framework: http://www.djangoproject.com/documentation/flatpages/
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.. _syndication framework: http://www.djangoproject.com/documentation/syndication/
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.. _authentication framework: http://www.djangoproject.com/documentation/authentication/
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