Django comes with an optional “sites” framework. It’s a hook for associating objects and functionality to particular Web sites, and it’s a holding place for the domain names and “verbose” names of your Django-powered sites.
Use it if your single Django installation powers more than one site and you need to differentiate between those sites in some way.
The whole sites framework is based on a simple model:
This model has domain and name fields. The SITE_ID setting specifies the database ID of the Site object associated with that particular settings file.
How you use this is up to you, but Django uses it in a couple of ways automatically via simple conventions.
Why would you use sites? It’s best explained through examples.
The Django-powered sites LJWorld.com and Lawrence.com are operated by the same news organization – the Lawrence Journal-World newspaper in Lawrence, Kansas. LJWorld.com focuses on news, while Lawrence.com focuses on local entertainment. But sometimes editors want to publish an article on both sites.
The brain-dead way of solving the problem would be to require site producers to publish the same story twice: once for LJWorld.com and again for Lawrence.com. But that’s inefficient for site producers, and it’s redundant to store multiple copies of the same story in the database.
The better solution is simple: Both sites use the same article database, and an article is associated with one or more sites. In Django model terminology, that’s represented by a ManyToManyField in the Article model:
from django.db import models
from django.contrib.sites.models import Site
class Article(models.Model):
headline = models.CharField(max_length=200)
# ...
sites = models.ManyToManyField(Site)
This accomplishes several things quite nicely:
It lets the site producers edit all content -- on both sites -- in a single interface (the Django admin).
It means the same story doesn't have to be published twice in the database; it only has a single record in the database.
It lets the site developers use the same Django view code for both sites. The view code that displays a given story just checks to make sure the requested story is on the current site. It looks something like this:
from django.conf import settings
def article_detail(request, article_id):
try:
a = Article.objects.get(id=article_id, sites__id__exact=settings.SITE_ID)
except Article.DoesNotExist:
raise Http404
# ...
Similarly, you can associate a model to the Site model in a many-to-one relationship, using ForeignKey.
For example, if an article is only allowed on a single site, you'd use a model like this:
from django.db import models
from django.contrib.sites.models import Site
class Article(models.Model):
headline = models.CharField(max_length=200)
# ...
site = models.ForeignKey(Site)
This has the same benefits as described in the last section.
On a lower level, you can use the sites framework in your Django views to do particular things based on the site in which the view is being called. For example:
from django.conf import settings
def my_view(request):
if settings.SITE_ID == 3:
# Do something.
else:
# Do something else.
Of course, it's ugly to hard-code the site IDs like that. This sort of hard-coding is best for hackish fixes that you need done quickly. A slightly cleaner way of accomplishing the same thing is to check the current site's domain:
from django.conf import settings
from django.contrib.sites.models import Site
def my_view(request):
current_site = Site.objects.get(id=settings.SITE_ID)
if current_site.domain == 'foo.com':
# Do something
else:
# Do something else.
The idiom of retrieving the Site object for the value of settings.SITE_ID is quite common, so the Site model's manager has a get_current() method. This example is equivalent to the previous one:
from django.contrib.sites.models import Site
def my_view(request):
current_site = Site.objects.get_current()
if current_site.domain == 'foo.com':
# Do something
else:
# Do something else.
LJWorld.com and Lawrence.com both have e-mail alert functionality, which lets readers sign up to get notifications when news happens. It's pretty basic: A reader signs up on a Web form, and he immediately gets an e-mail saying, "Thanks for your subscription."
It'd be inefficient and redundant to implement this signup-processing code twice, so the sites use the same code behind the scenes. But the "thank you for signing up" notice needs to be different for each site. By using Site objects, we can abstract the "thank you" notice to use the values of the current site's name and domain.
Here's an example of what the form-handling view looks like:
from django.contrib.sites.models import Site
from django.core.mail import send_mail
def register_for_newsletter(request):
# Check form values, etc., and subscribe the user.
# ...
current_site = Site.objects.get_current()
send_mail('Thanks for subscribing to %s alerts' % current_site.name,
'Thanks for your subscription. We appreciate it.\n\n-The %s team.' % current_site.name,
'editor@%s' % current_site.domain,
[user.email])
# ...
On Lawrence.com, this e-mail has the subject line "Thanks for subscribing to lawrence.com alerts." On LJWorld.com, the e-mail has the subject "Thanks for subscribing to LJWorld.com alerts." Same goes for the e-mail's message body.
Note that an even more flexible (but more heavyweight) way of doing this would be to use Django's template system. Assuming Lawrence.com and LJWorld.com have different template directories (TEMPLATE_DIRS), you could simply farm out to the template system like so:
from django.core.mail import send_mail
from django.template import loader, Context
def register_for_newsletter(request):
# Check form values, etc., and subscribe the user.
# ...
subject = loader.get_template('alerts/subject.txt').render(Context({}))
message = loader.get_template('alerts/message.txt').render(Context({}))
send_mail(subject, message, 'editor@ljworld.com', [user.email])
# ...
In this case, you'd have to create subject.txt and message.txt template files for both the LJWorld.com and Lawrence.com template directories. That gives you more flexibility, but it's also more complex.
It's a good idea to exploit the Site objects as much as possible, to remove unneeded complexity and redundancy.
Django's get_absolute_url() convention is nice for getting your objects' URL without the domain name, but in some cases you might want to display the full URL -- with http:// and the domain and everything -- for an object. To do this, you can use the sites framework. A simple example:
>>> from django.contrib.sites.models import Site
>>> obj = MyModel.objects.get(id=3)
>>> obj.get_absolute_url()
'/mymodel/objects/3/'
>>> Site.objects.get_current().domain
'example.com'
>>> 'http://%s%s' % (Site.objects.get_current().domain, obj.get_absolute_url())
'http://example.com/mymodel/objects/3/'
As the current site is stored in the database, each call to Site.objects.get_current() could result in a database query. But Django is a little cleverer than that: on the first request, the current site is cached, and any subsequent call returns the cached data instead of hitting the database.
If for any reason you want to force a database query, you can tell Django to clear the cache using Site.objects.clear_cache():
# First call; current site fetched from database.
current_site = Site.objects.get_current()
# ...
# Second call; current site fetched from cache.
current_site = Site.objects.get_current()
# ...
# Force a database query for the third call.
Site.objects.clear_cache()
current_site = Site.objects.get_current()
If Sites play a key role in your application, consider using the helpful CurrentSiteManager in your model(s). It's a model manager that automatically filters its queries to include only objects associated with the current Site.
Use CurrentSiteManager by adding it to your model explicitly. For example:
from django.db import models
from django.contrib.sites.models import Site
from django.contrib.sites.managers import CurrentSiteManager
class Photo(models.Model):
photo = models.FileField(upload_to='/home/photos')
photographer_name = models.CharField(max_length=100)
pub_date = models.DateField()
site = models.ForeignKey(Site)
objects = models.Manager()
on_site = CurrentSiteManager()
With this model, Photo.objects.all() will return all Photo objects in the database, but Photo.on_site.all() will return only the Photo objects associated with the current site, according to the SITE_ID setting.
Put another way, these two statements are equivalent:
Photo.objects.filter(site=settings.SITE_ID)
Photo.on_site.all()
How did CurrentSiteManager know which field of Photo was the Site? It defaults to looking for a field called Site. If your model has a ForeignKey or ManyToManyField called something other than Site, you need to explicitly pass that as the parameter to CurrentSiteManager. The following model, which has a field called publish_on, demonstrates this:
from django.db import models
from django.contrib.sites.models import Site
from django.contrib.sites.managers import CurrentSiteManager
class Photo(models.Model):
photo = models.FileField(upload_to='/home/photos')
photographer_name = models.CharField(max_length=100)
pub_date = models.DateField()
publish_on = models.ForeignKey(Site)
objects = models.Manager()
on_site = CurrentSiteManager('publish_on')
If you attempt to use CurrentSiteManager and pass a field name that doesn't exist, Django will raise a ValueError.
Finally, note that you'll probably want to keep a normal (non-site-specific) Manager on your model, even if you use CurrentSiteManager. As explained in the manager documentation, if you define a manager manually, then Django won't create the automatic objects = models.Manager() manager for you.Also, note that certain parts of Django -- namely, the Django admin site and generic views -- use whichever manager is defined first in the model, so if you want your admin site to have access to all objects (not just site-specific ones), put objects = models.Manager() in your model, before you define CurrentSiteManager.
Although it's not required that you use the sites framework, it's strongly encouraged, because Django takes advantage of it in a few places. Even if your Django installation is powering only a single site, you should take the two seconds to create the site object with your domain and name, and point to its ID in your SITE_ID setting.
Here's how Django uses the sites framework:
Some django.contrib applications take advantage of the sites framework but are architected in a way that doesn't require the sites framework to be installed in your database. (Some people don't want to, or just aren't able to install the extra database table that the sites framework requires.) For those cases, the framework provides a RequestSite class, which can be used as a fallback when the database-backed sites framework is not available.
A RequestSite object has a similar interface to a normal Site object, except its __init__() method takes an HttpRequest object. It's able to deduce the domain and name by looking at the request's domain. It has save() and delete() methods to match the interface of Site, but the methods raise NotImplementedError.
Sep 20, 2009