402 lines
16 KiB
Plaintext
402 lines
16 KiB
Plaintext
.. _ref-contrib-sites:
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=====================
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The "sites" framework
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=====================
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.. module:: django.contrib.sites
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:synopsis: Lets you operate multiple web sites from the same database and
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Django project
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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 a simple model:
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.. class:: django.contrib.sites.models.Site
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This model has :attr:`~django.contrib.sites.models.Site.domain` and
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:attr:`~django.contrib.sites.models.Site.name` fields. The :setting:`SITE_ID`
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setting specifies the database ID of the
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:class:`~django.contrib.sites.models.Site` object associated with that
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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 :class:`~django.db.models.ManyToManyField` in the
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``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(max_length=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 :class:`~django.contrib.sites.models.Site`
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model in a many-to-one relationship, using
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:class:`~django.db.models.fields.related.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(max_length=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 the 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 :class:`~django.contrib.sites.models.Site` object
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for the value of :setting:`settings.SITE_ID <SITE_ID>` is quite common, so
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the :class:`~django.contrib.sites.models.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
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:class:`~django.contrib.sites.models.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 :attr:`~django.contrib.sites.models.Site.name` and
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:attr:`~django.contrib.sites.models.Site.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 (:setting:`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 :file:`subject.txt` and :file:`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 :class:`~django.contrib.sites.models.Site``
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objects as much as possible, to remove 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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Caching the current ``Site`` object
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===================================
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**New in Django development version**
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As the current site is stored in the database, each call to
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``Site.objects.get_current()`` could result in a database query. But Django is a
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little cleverer than that: on the first request, the current site is cached, and
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any subsequent call returns the cached data instead of hitting the database.
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If for any reason you want to force a database query, you can tell Django to
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clear the cache using ``Site.objects.clear_cache()``::
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# First call; current site fetched from database.
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current_site = Site.objects.get_current()
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# ...
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# Second call; current site fetched from cache.
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current_site = Site.objects.get_current()
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# ...
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# Force a database query for the third call.
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Site.objects.clear_cache()
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current_site = Site.objects.get_current()
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The ``CurrentSiteManager``
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==========================
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.. class:: django.contrib.sites.managers.CurrentSiteManager
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If :class:`~django.contrib.sites.models.Site`\s play a key role in your application,
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consider using the helpful
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:class:`~django.contrib.sites.managers.CurrentSiteManager` in your model(s).
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It's a model :ref:`manager <topics-db-managers>` that automatically filters
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its queries to include only objects associated with the current
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:class:`~django.contrib.sites.models.Site`.
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Use :class:`~django.contrib.sites.managers.CurrentSiteManager` by adding it to
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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(max_length=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`` objects
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associated with the current site, according to the :setting:`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 :class:`~django.contrib.sites.managers.CurrentSiteManager` know which
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field of ``Photo`` was the :class:`~django.contrib.sites.models.Site`? It
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defaults to looking for a field called
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:class:`~django.contrib.sites.models.Site`. If your model has a
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:class:`~django.db.models.fields.related.ForeignKey` or
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:class:`~django.db.models.fields.related.ManyToManyField` called something
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*other* than :class:`~django.contrib.sites.models.Site`, you need to explicitly
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pass that as the parameter to
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:class:`~django.contrib.sites.managers.CurrentSiteManager`. The following model,
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which has a field called ``publish_on``, demonstrates 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(max_length=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 :class:`~django.contrib.sites.managers.CurrentSiteManager`
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and pass a field name that doesn't exist, Django will raise a :exc:`ValueError`.
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Finally, note that you'll probably want to keep a normal (non-site-specific)
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``Manager`` on your model, even if you use
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:class:`~django.contrib.sites.managers.CurrentSiteManager`. As explained
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in the :ref:`manager documentation <topics-db-managers>`, if you define a manager
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manually, then Django won't create the automatic ``objects = models.Manager()``
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manager for you.Also, note that certain parts of Django -- namely, the Django admin site and
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generic views -- use whichever manager is defined *first* in the model, so if
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you want your admin site to have access to all objects (not just site-specific
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ones), put ``objects = models.Manager()`` in your model, before you define
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:class:`~django.contrib.sites.managers.CurrentSiteManager`.
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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 :setting:`SITE_ID` setting.
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Here's how Django uses the sites framework:
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* In the :mod:`redirects framework <django.contrib.redirects>`, each
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redirect object is associated with a particular site. When Django searches
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for a redirect, it takes into account the current :setting:`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
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:class:`~django.contrib.sites.models.Site` is set to the current
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:setting:`SITE_ID`, and when comments are listed via the appropriate
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template tag, only the comments for the current site are displayed.
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* In the :mod:`flatpages framework <django.contrib.flatpages>`, each
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flatpage is associated with a particular site. When a flatpage is created,
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you specify its :class:`~django.contrib.sites.models.Site`, and the
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:class:`~django.contrib.flatpages.middleware.FlatpageFallbackMiddleware`
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checks the current :setting:`SITE_ID` in retrieving flatpages to display.
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* In the :mod:`syndication framework <django.contrib.syndication>`, the
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templates for ``title`` and ``description`` automatically have access to a
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variable ``{{ site }}``, which is the
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:class:`~django.contrib.sites.models.Site` object representing the current
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site. Also, the hook for providing item URLs will use the ``domain`` from
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the current :class:`~django.contrib.sites.models.Site` object if you don't
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specify a fully-qualified domain.
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* In the :mod:`authentication framework <django.contrib.auth>`, the
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:func:`django.contrib.auth.views.login` view passes the current
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:class:`~django.contrib.sites.models.Site` name to the template as
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``{{ site_name }}``.
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* The shortcut view (:func:`django.views.defaults.shortcut`) uses the domain
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of the current :class:`~django.contrib.sites.models.Site` object when
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calculating an object's URL.
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* In the admin framework, the "view on site" link uses the current
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:class:`~django.contrib.sites.models.Site` to work out the domain for the
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site that it will redirect to.
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``RequestSite`` objects
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=======================
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.. _requestsite-objects:
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**New in Django development version**
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Some :ref:`django.contrib <ref-contrib-index>` applications take advantage of
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the sites framework but are architected in a way that doesn't *require* the
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sites framework to be installed in your database. (Some people don't want to, or
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just aren't *able* to install the extra database table that the sites framework
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requires.) For those cases, the framework provides a
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:class:`~django.contrib.sites.models.RequestSite` class, which can be used as a
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fallback when the database-backed sites framework is not available.
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A :class:`~django.contrib.sites.models.RequestSite` object has a similar
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interface to a normal :class:`~django.contrib.sites.models.Site` object, except
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its :meth:`~django.contrib.sites.models.RequestSite.__init__()` method takes an
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:class:`~django.http.HttpRequest` object. It's able to deduce the
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:attr:`~django.contrib.sites.models.RequestSite.domain` and
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:attr:`~django.contrib.sites.models.RequestSite.name` by looking at the
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request's domain. It has :meth:`~django.contrib.sites.models.RequestSite.save()`
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and :meth:`~django.contrib.sites.models.RequestSite.delete()` methods to match
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the interface of :class:`~django.contrib.sites.models.Site`, but the methods
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raise :exc:`NotImplementedError`.
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