Django comes with an optional "sites" framework. It's a hook for associating objects and functionality to particular websites, 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 sites framework is mainly based on this model:
models.Site¶A model for storing the domain and name attributes of a website.
domain¶The fully qualified domain name associated with the website.
For example, www.example.com.
name¶A human-readable "verbose" name for the website.
The SITE_ID setting specifies the database ID of the
Site object associated with that
particular settings file. If the setting is omitted, the
get_current_site() function will
try to get the current site by comparing the
domain with the host name from
the request.get_host() method.
How you use this is up to you, but Django uses it in a couple of ways automatically via a couple of conventions.
Why would you use sites? It's best explained through examples.
The LJWorld.com and Lawrence.com sites are operated by the same news organization -- the Lawrence Journal-World newspaper in Lawrence, Kansas. LJWorld.com focused on news, while Lawrence.com focused on local entertainment. But sometimes editors wanted to publish an article on both sites.
The naive 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.
A better solution removes the content duplication: 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.contrib.sites.models import Site
from django.db import models
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 checks to make sure the requested story is on the current site. It looks something like this:
from django.contrib.sites.shortcuts import get_current_site
def article_detail(request, article_id):
try:
a = Article.objects.get(id=article_id, sites__id=get_current_site(request).id)
except Article.DoesNotExist:
raise Http404("Article does not exist on this 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.contrib.sites.models import Site
from django.db import models
class Article(models.Model):
headline = models.CharField(max_length=200)
# ...
site = models.ForeignKey(Site, on_delete=models.CASCADE)
This has the same benefits as described in the last section.
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.
pass
else:
# Do something else.
pass
It's fragile to hard-code the site IDs like that, in case they change. The cleaner way of accomplishing the same thing is to check the current site's domain:
from django.contrib.sites.shortcuts import get_current_site
def my_view(request):
current_site = get_current_site(request)
if current_site.domain == 'foo.com':
# Do something
pass
else:
# Do something else.
pass
This has also the advantage of checking if the sites framework is installed,
and return a RequestSite instance if
it is not.
If you don't have access to the request object, you can use the
get_current() method of the Site
model's manager. You should then ensure that your settings file does contain
the SITE_ID setting. This example is equivalent to the previous one:
from django.contrib.sites.models import Site
def my_function_without_request():
current_site = Site.objects.get_current()
if current_site.domain == 'foo.com':
# Do something
pass
else:
# Do something else.
pass
LJWorld.com and Lawrence.com both have email 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 immediately gets an email saying, "Thanks for your subscription."
It'd be inefficient and redundant to implement this sign up 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.shortcuts import get_current_site
from django.core.mail import send_mail
def register_for_newsletter(request):
# Check form values, etc., and subscribe the user.
# ...
current_site = get_current_site(request)
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 email has the subject line "Thanks for subscribing to lawrence.com alerts." On LJWorld.com, the email has the subject "Thanks for subscribing to LJWorld.com alerts." Same goes for the email'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 (DIRS), you could
farm out to the template system like so:
from django.core.mail import send_mail
from django.template import loader
def register_for_newsletter(request):
# Check form values, etc., and subscribe the user.
# ...
subject = loader.get_template('alerts/subject.txt').render({})
message = loader.get_template('alerts/message.txt').render({})
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. An 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'
>>> 'https://%s%s' % (Site.objects.get_current().domain, obj.get_absolute_url())
'https://example.com/mymodel/objects/3/'
To enable the sites framework, follow these steps:
Add 'django.contrib.sites' to your INSTALLED_APPS setting.
Define a SITE_ID setting:
SITE_ID = 1
Run migrate.
django.contrib.sites registers a
post_migrate signal handler which creates a
default site named example.com with the domain example.com. This site
will also be created after Django creates the test database. To set the
correct name and domain for your project, you can use a data migration.
In order to serve different sites in production, you'd create a separate
settings file with each SITE_ID (perhaps importing from a common settings
file to avoid duplicating shared settings) and then specify the appropriate
DJANGO_SETTINGS_MODULE for each site.
Site object¶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()
CurrentSiteManager¶managers.CurrentSiteManager¶If Site plays 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.
Mandatory SITE_ID
The CurrentSiteManager is only usable when the SITE_ID
setting is defined in your settings.
Use CurrentSiteManager by adding it to
your model explicitly. For example:
from django.contrib.sites.models import Site
from django.contrib.sites.managers import CurrentSiteManager
from django.db import models
class Photo(models.Model):
photo = models.FileField(upload_to='photos')
photographer_name = models.CharField(max_length=100)
pub_date = models.DateField()
site = models.ForeignKey(Site, on_delete=models.CASCADE)
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? By default,
CurrentSiteManager looks for a
either a ForeignKey called
site or a
ManyToManyField called
sites to filter on. If you use a field named something other than
site or sites to identify which
Site objects your object is
related to, then you need to explicitly pass the custom field name as
a parameter to
CurrentSiteManager on your
model. The following model, which has a field called publish_on,
demonstrates this:
from django.contrib.sites.models import Site
from django.contrib.sites.managers import CurrentSiteManager
from django.db import models
class Photo(models.Model):
photo = models.FileField(upload_to='photos')
photographer_name = models.CharField(max_length=100)
pub_date = models.DateField()
publish_on = models.ForeignKey(Site, on_delete=models.CASCADE)
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.
If you often use this pattern:
from django.contrib.sites.models import Site
def my_view(request):
site = Site.objects.get_current()
...
To avoid repetitions, add
django.contrib.sites.middleware.CurrentSiteMiddleware to
MIDDLEWARE. The middleware sets the site attribute on every
request object, so you can use request.site to get the current site.
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:
redirects framework, each
redirect object is associated with a particular site. When Django searches
for a redirect, it takes into account the current site.flatpages framework, each
flatpage is associated with a particular site. When a flatpage is created,
you specify its Site, and the
FlatpageFallbackMiddleware
checks the current site in retrieving flatpages to display.syndication framework, the
templates for title and description automatically have access to a
variable {{ site }}, which is the
Site object representing the current
site. Also, the hook for providing item URLs will use the domain from
the current Site object if you don't
specify a fully-qualified domain.authentication framework,
django.contrib.auth.views.LoginView passes the current
Site name to the template as
{{ site_name }}.django.contrib.contenttypes.views.shortcut)
uses the domain of the current
Site object when calculating
an object's URL.Site to work out the domain for the
site that it will redirect to.RequestSite objects¶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
django.contrib.sites.requests.RequestSite class, which can be used as
a fallback when the database-backed sites framework is not available.
requests.RequestSite¶A class that shares the primary interface of
Site (i.e., it has
domain and name attributes) but gets its data from a Django
HttpRequest object rather than from a database.
__init__(request)¶Sets the name and domain attributes to the value of
get_host().
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.
get_current_site shortcut¶Finally, to avoid repetitive fallback code, the framework provides a
django.contrib.sites.shortcuts.get_current_site() function.
shortcuts.get_current_site(request)¶A function that checks if django.contrib.sites is installed and
returns either the current Site
object or a RequestSite object
based on the request. It looks up the current site based on
request.get_host() if the
SITE_ID setting is not defined.
Both a domain and a port may be returned by request.get_host() when the Host header has a port
explicitly specified, e.g. example.com:80. In such cases, if the
lookup fails because the host does not match a record in the database,
the port is stripped and the lookup is retried with the domain part
only. This does not apply to
RequestSite which will always
use the unmodified host.
2022年6月01日