Associating content with multiple sites

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(maxlength=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
      # ...
  

Associating content with a single site

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(maxlength=200)
    # ...
    site = models.ForeignKey(Site)

This has the same benefits as described in the last section.

Hooking into the current site from views

On a lower level, you can use the sites framework in your Django views to do particular things based on what 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.

Getting the current domain for display

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.

Getting the current domain for full URLs

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/'