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518 lines
20 KiB
Plaintext
=====================
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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 websites 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 websites, 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 sites framework is mainly based on this model:
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.. class:: models.Site
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A model for storing the ``domain`` and ``name`` attributes of a website.
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.. attribute:: domain
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The fully qualified domain name associated with the website.
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For example, ``www.example.com``.
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.. attribute:: name
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A human-readable "verbose" name for the website.
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The :setting:`SITE_ID` 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. If the setting is omitted, the
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:func:`~django.contrib.sites.shortcuts.get_current_site` function will
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try to get the current site by comparing the
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:attr:`~django.contrib.sites.models.Site.domain` with the host name from
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the :meth:`request.get_host() <django.http.HttpRequest.get_host>` method.
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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 a couple of 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 LJWorld.com_ and Lawrence.com sites were operated by the same news
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organization -- the Lawrence Journal-World newspaper in Lawrence, Kansas.
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LJWorld.com focused on news, while Lawrence.com focused on local entertainment.
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But sometimes editors wanted to publish an article on *both* sites.
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The naive 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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A better solution removes the content duplication: Both sites use the same
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article database, and an article is associated with one or more sites. In
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Django model terminology, that's represented by a
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:class:`~django.db.models.ManyToManyField` in the ``Article`` model::
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from django.contrib.sites.models import Site
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from django.db import models
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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 checks to make sure the requested
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story is on the current site. It looks something like this::
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from django.contrib.sites.shortcuts import get_current_site
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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=get_current_site(request).id)
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except Article.DoesNotExist:
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raise Http404("Article does not exist on this site")
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# ...
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.. _ljworld.com: https://www2.ljworld.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
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: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.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.contrib.sites.models import Site
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from django.db import models
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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, on_delete=models.CASCADE)
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This has the same benefits as described in the last section.
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.. _hooking-into-current-site-from-views:
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Hooking into the current site from views
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----------------------------------------
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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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pass
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else:
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# Do something else.
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pass
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It's fragile to hard-code the site IDs like that, in case they change. The
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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.contrib.sites.shortcuts import get_current_site
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def my_view(request):
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current_site = get_current_site(request)
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if current_site.domain == "foo.com":
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# Do something
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pass
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else:
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# Do something else.
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pass
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This has also the advantage of checking if the sites framework is installed,
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and return a :class:`~django.contrib.sites.requests.RequestSite` instance if
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it is not.
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If you don't have access to the request object, you can use the
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``get_current()`` method of the :class:`~django.contrib.sites.models.Site`
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model's manager. You should then ensure that your settings file does contain
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the :setting:`SITE_ID` setting. 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_function_without_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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pass
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else:
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# Do something else.
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pass
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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 email 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 immediately gets an email saying,
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"Thanks for your subscription."
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It'd be inefficient and redundant to implement this sign up 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.shortcuts import get_current_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 = get_current_site(request)
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send_mail(
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"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."
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% (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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# ...
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On Lawrence.com, this email has the subject line "Thanks for subscribing to
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lawrence.com alerts." On LJWorld.com, the email has the subject "Thanks for
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subscribing to LJWorld.com alerts." Same goes for the email'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:`DIRS <TEMPLATES-DIRS>`), you could
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farm out 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
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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({})
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message = loader.get_template("alerts/message.txt").render({})
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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`
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template files for both the LJWorld.com and Lawrence.com template directories.
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That 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. An example:
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.. code-block:: pycon
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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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>>> "https://%s%s" % (Site.objects.get_current().domain, obj.get_absolute_url())
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'https://example.com/mymodel/objects/3/'
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.. _enabling-the-sites-framework:
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Enabling the sites framework
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============================
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To enable the sites framework, follow these steps:
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#. Add ``'django.contrib.sites'`` to your :setting:`INSTALLED_APPS` setting.
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#. Define a :setting:`SITE_ID` setting::
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SITE_ID = 1
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#. Run :djadmin:`migrate`.
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``django.contrib.sites`` registers a
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:data:`~django.db.models.signals.post_migrate` signal handler which creates a
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default site named ``example.com`` with the domain ``example.com``. This site
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will also be created after Django creates the test database. To set the
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correct name and domain for your project, you can use a :ref:`data migration
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<data-migrations>`.
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In order to serve different sites in production, you'd create a separate
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settings file with each ``SITE_ID`` (perhaps importing from a common settings
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file to avoid duplicating shared settings) and then specify the appropriate
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:envvar:`DJANGO_SETTINGS_MODULE` for each site.
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Caching the current ``Site`` object
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===================================
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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:: managers.CurrentSiteManager
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If :class:`~django.contrib.sites.models.Site` plays a key role in your
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application, consider using the helpful
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:class:`~django.contrib.sites.managers.CurrentSiteManager` in your
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model(s). It's a model :doc:`manager </topics/db/managers>` that
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automatically filters its queries to include only objects associated
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with the current :class:`~django.contrib.sites.models.Site`.
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.. admonition:: Mandatory :setting:`SITE_ID`
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The ``CurrentSiteManager`` is only usable when the :setting:`SITE_ID`
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setting is defined in your settings.
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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.contrib.sites.models import Site
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from django.contrib.sites.managers import CurrentSiteManager
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from django.db import models
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class Photo(models.Model):
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photo = models.FileField(upload_to="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, on_delete=models.CASCADE)
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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`
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know which field of ``Photo`` was the
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:class:`~django.contrib.sites.models.Site`? By default,
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:class:`~django.contrib.sites.managers.CurrentSiteManager` looks for a
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either a :class:`~django.db.models.ForeignKey` called
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``site`` or a
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:class:`~django.db.models.ManyToManyField` called
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``sites`` to filter on. If you use a field named something other than
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``site`` or ``sites`` to identify which
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:class:`~django.contrib.sites.models.Site` objects your object is
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related to, then you need to explicitly pass the custom field name as
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a parameter to
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:class:`~django.contrib.sites.managers.CurrentSiteManager` on your
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model. The following model, which has a field called ``publish_on``,
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demonstrates this::
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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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from django.db import models
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class Photo(models.Model):
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photo = models.FileField(upload_to="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, on_delete=models.CASCADE)
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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 ``ValueError``.
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Finally, note that you'll probably want to keep a normal
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(non-site-specific) ``Manager`` on your model, even if you use
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:class:`~django.contrib.sites.managers.CurrentSiteManager`. As
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explained in the :doc:`manager documentation </topics/db/managers>`, if
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you define a manager manually, then Django won't create the automatic
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``objects = models.Manager()`` manager for you. Also note that certain
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parts of Django -- namely, the Django admin site and generic views --
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use whichever manager is defined *first* in the model, so if you want
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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
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define :class:`~django.contrib.sites.managers.CurrentSiteManager`.
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.. _site-middleware:
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Site middleware
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===============
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If you often use this pattern::
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from django.contrib.sites.models import Site
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def my_view(request):
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site = Site.objects.get_current()
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...
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To avoid repetitions, add
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:class:`django.contrib.sites.middleware.CurrentSiteMiddleware` to
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:setting:`MIDDLEWARE`. The middleware sets the ``site`` attribute on every
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request object, so you can use ``request.site`` to get the current site.
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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 site.
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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 site 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>`,
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:class:`django.contrib.auth.views.LoginView` 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 (``django.contrib.contenttypes.views.shortcut``)
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uses the domain of the current
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:class:`~django.contrib.sites.models.Site` object when calculating
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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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Some :doc:`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,
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or just aren't *able* to install the extra database table that the sites
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framework requires.) For those cases, the framework provides a
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:class:`django.contrib.sites.requests.RequestSite` class, which can be used as
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a fallback when the database-backed sites framework is not available.
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.. class:: requests.RequestSite
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A class that shares the primary interface of
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:class:`~django.contrib.sites.models.Site` (i.e., it has
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``domain`` and ``name`` attributes) but gets its data from a Django
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:class:`~django.http.HttpRequest` object rather than from a database.
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.. method:: __init__(request)
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Sets the ``name`` and ``domain`` attributes to the value of
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:meth:`~django.http.HttpRequest.get_host`.
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A :class:`~django.contrib.sites.requests.RequestSite` object has a similar
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interface to a normal :class:`~django.contrib.sites.models.Site` object,
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except its :meth:`~django.contrib.sites.requests.RequestSite.__init__()`
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method takes an :class:`~django.http.HttpRequest` object. It's able to deduce
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the ``domain`` and ``name`` by looking at the request's domain. It has
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``save()`` and ``delete()`` methods to match the interface of
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:class:`~django.contrib.sites.models.Site`, but the methods raise
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:exc:`NotImplementedError`.
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``get_current_site`` shortcut
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=============================
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Finally, to avoid repetitive fallback code, the framework provides a
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:func:`django.contrib.sites.shortcuts.get_current_site` function.
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.. function:: shortcuts.get_current_site(request)
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A function that checks if ``django.contrib.sites`` is installed and
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returns either the current :class:`~django.contrib.sites.models.Site`
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object or a :class:`~django.contrib.sites.requests.RequestSite` object
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based on the request. It looks up the current site based on
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:meth:`request.get_host() <django.http.HttpRequest.get_host>` if the
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:setting:`SITE_ID` setting is not defined.
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Both a domain and a port may be returned by :meth:`request.get_host()
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<django.http.HttpRequest.get_host>` when the Host header has a port
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explicitly specified, e.g. ``example.com:80``. In such cases, if the
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lookup fails because the host does not match a record in the database,
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the port is stripped and the lookup is retried with the domain part
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only. This does not apply to
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:class:`~django.contrib.sites.requests.RequestSite` which will always
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use the unmodified host.
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