Django comes with an optional "sites" framework. It's a hook for associating\nobjects and functionality to particular websites, and it's a holding place for\nthe domain names and "verbose" names of your Django-powered sites.
\nUse it if your single Django installation powers more than one site and you\nneed to differentiate between those sites in some way.
\nThe sites framework is mainly based on this model:
\nA model for storing the domain and name attributes of a website.
The fully qualified domain name associated with the website.\nFor example, www.example.com.
A human-readable "verbose" name for the website.
\nThe SITE_ID setting specifies the database ID of the\nSite object associated with that\nparticular settings file. If the setting is omitted, the\nget_current_site() function will\ntry to get the current site by comparing the\ndomain with the host name from\nthe request.get_host() method.
How you use this is up to you, but Django uses it in a couple of ways\nautomatically via a couple of conventions.
\nWhy would you use sites? It's best explained through examples.
\nThe LJWorld.com and Lawrence.com sites are operated by the same news\norganization -- the Lawrence Journal-World newspaper in Lawrence, Kansas.\nLJWorld.com focused on news, while Lawrence.com focused on local entertainment.\nBut sometimes editors wanted to publish an article on both sites.
\nThe naive way of solving the problem would be to require site producers to\npublish the same story twice: once for LJWorld.com and again for Lawrence.com.\nBut that's inefficient for site producers, and it's redundant to store\nmultiple copies of the same story in the database.
\nA better solution removes the content duplication: Both sites use the same\narticle database, and an article is associated with one or more sites. In\nDjango model terminology, that's represented by a\nManyToManyField in the Article model:
from django.contrib.sites.models import Site\nfrom django.db import models\n\nclass Article(models.Model):\n headline = models.CharField(max_length=200)\n # ...\n sites = models.ManyToManyField(Site)\nThis accomplishes several things quite nicely:
\nIt lets the site producers edit all content -- on both sites -- in a\nsingle interface (the Django admin).
It means the same story doesn't have to be published twice in the\ndatabase; it only has a single record in the database.
It lets the site developers use the same Django view code for both sites.\nThe view code that displays a given story checks to make sure the requested\nstory is on the current site. It looks something like this:
\nfrom django.contrib.sites.shortcuts import get_current_site\n\ndef article_detail(request, article_id):\n try:\n a = Article.objects.get(id=article_id, sites__id=get_current_site(request).id)\n except Article.DoesNotExist:\n raise Http404("Article does not exist on this site")\n # ...\nSimilarly, you can associate a model to the\nSite\nmodel in a many-to-one relationship, using\nForeignKey.
For example, if an article is only allowed on a single site, you'd use a model\nlike this:
\nfrom django.contrib.sites.models import Site\nfrom django.db import models\n\nclass Article(models.Model):\n headline = models.CharField(max_length=200)\n # ...\n site = models.ForeignKey(Site, on_delete=models.CASCADE)\nThis has the same benefits as described in the last section.
\nYou can use the sites framework in your Django views to do\nparticular things based on the site in which the view is being called.\nFor example:
\nfrom django.conf import settings\n\ndef my_view(request):\n if settings.SITE_ID == 3:\n # Do something.\n pass\n else:\n # Do something else.\n pass\nIt's fragile to hard-code the site IDs like that, in case they change. The\ncleaner way of accomplishing the same thing is to check the current site's\ndomain:
\nfrom django.contrib.sites.shortcuts import get_current_site\n\ndef my_view(request):\n current_site = get_current_site(request)\n if current_site.domain == 'foo.com':\n # Do something\n pass\n else:\n # Do something else.\n pass\nThis has also the advantage of checking if the sites framework is installed,\nand return a RequestSite instance if\nit is not.
If you don't have access to the request object, you can use the\nget_current() method of the Site\nmodel's manager. You should then ensure that your settings file does contain\nthe SITE_ID setting. This example is equivalent to the previous one:
from django.contrib.sites.models import Site\n\ndef my_function_without_request():\n current_site = Site.objects.get_current()\n if current_site.domain == 'foo.com':\n # Do something\n pass\n else:\n # Do something else.\n pass\nLJWorld.com and Lawrence.com both have email alert functionality, which lets\nreaders sign up to get notifications when news happens. It's pretty basic: A\nreader signs up on a Web form and immediately gets an email saying,\n"Thanks for your subscription."
\nIt'd be inefficient and redundant to implement this sign up processing code\ntwice, so the sites use the same code behind the scenes. But the "thank you for\nsigning up" notice needs to be different for each site. By using\nSite\nobjects, we can abstract the "thank you" notice to use the values of the\ncurrent site's name and\ndomain.
Here's an example of what the form-handling view looks like:
\nfrom django.contrib.sites.shortcuts import get_current_site\nfrom django.core.mail import send_mail\n\ndef register_for_newsletter(request):\n # Check form values, etc., and subscribe the user.\n # ...\n\n current_site = get_current_site(request)\n send_mail(\n 'Thanks for subscribing to %s alerts' % current_site.name,\n 'Thanks for your subscription. We appreciate it.\\n\\n-The %s team.' % (\n current_site.name,\n ),\n 'editor@%s' % current_site.domain,\n [user.email],\n )\n\n # ...\nOn Lawrence.com, this email has the subject line "Thanks for subscribing to\nlawrence.com alerts." On LJWorld.com, the email has the subject "Thanks for\nsubscribing to LJWorld.com alerts." Same goes for the email's message body.
\nNote that an even more flexible (but more heavyweight) way of doing this would\nbe to use Django's template system. Assuming Lawrence.com and LJWorld.com have\ndifferent template directories (DIRS), you could\nfarm out to the template system like so:
from django.core.mail import send_mail\nfrom django.template import loader\n\ndef register_for_newsletter(request):\n # Check form values, etc., and subscribe the user.\n # ...\n\n subject = loader.get_template('alerts/subject.txt').render({})\n message = loader.get_template('alerts/message.txt').render({})\n send_mail(subject, message, 'editor@ljworld.com', [user.email])\n\n # ...\nIn this case, you'd have to create subject.txt and message.txt\ntemplate files for both the LJWorld.com and Lawrence.com template directories.\nThat gives you more flexibility, but it's also more complex.
It's a good idea to exploit the Site\nobjects as much as possible, to remove unneeded complexity and redundancy.
Django's get_absolute_url() convention is nice for getting your objects'\nURL without the domain name, but in some cases you might want to display the\nfull URL -- with http:// and the domain and everything -- for an object.\nTo do this, you can use the sites framework. An example:
>>> from django.contrib.sites.models import Site\n>>> obj = MyModel.objects.get(id=3)\n>>> obj.get_absolute_url()\n'/mymodel/objects/3/'\n>>> Site.objects.get_current().domain\n'example.com'\n>>> 'https://%s%s' % (Site.objects.get_current().domain, obj.get_absolute_url())\n'https://example.com/mymodel/objects/3/'\nTo enable the sites framework, follow these steps:
\nAdd 'django.contrib.sites' to your INSTALLED_APPS setting.
Define a SITE_ID setting:
SITE_ID = 1\nRun migrate.
django.contrib.sites registers a\npost_migrate signal handler which creates a\ndefault site named example.com with the domain example.com. This site\nwill also be created after Django creates the test database. To set the\ncorrect 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\nsettings file with each SITE_ID (perhaps importing from a common settings\nfile to avoid duplicating shared settings) and then specify the appropriate\nDJANGO_SETTINGS_MODULE for each site.
Site objectLink to this heading#As the current site is stored in the database, each call to\nSite.objects.get_current() could result in a database query. But Django is a\nlittle cleverer than that: on the first request, the current site is cached, and\nany 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\nclear the cache using Site.objects.clear_cache():
# First call; current site fetched from database.\ncurrent_site = Site.objects.get_current()\n# ...\n\n# Second call; current site fetched from cache.\ncurrent_site = Site.objects.get_current()\n# ...\n\n# Force a database query for the third call.\nSite.objects.clear_cache()\ncurrent_site = Site.objects.get_current()\nCurrentSiteManagerLink to this heading#If Site plays a key role in your\napplication, consider using the helpful\nCurrentSiteManager in your\nmodel(s). It's a model manager that\nautomatically filters its queries to include only objects associated\nwith the current Site.
Use CurrentSiteManager by adding it to\nyour model explicitly. For example:
from django.contrib.sites.models import Site\nfrom django.contrib.sites.managers import CurrentSiteManager\nfrom django.db import models\n\nclass Photo(models.Model):\n photo = models.FileField(upload_to='photos')\n photographer_name = models.CharField(max_length=100)\n pub_date = models.DateField()\n site = models.ForeignKey(Site, on_delete=models.CASCADE)\n objects = models.Manager()\n on_site = CurrentSiteManager()\nWith this model, Photo.objects.all() will return all Photo objects in\nthe database, but Photo.on_site.all() will return only the Photo objects\nassociated with the current site, according to the SITE_ID setting.
Put another way, these two statements are equivalent:
\nPhoto.objects.filter(site=settings.SITE_ID)\nPhoto.on_site.all()\nHow did CurrentSiteManager\nknow which field of Photo was the\nSite? By default,\nCurrentSiteManager looks for a\neither a ForeignKey called\nsite or a\nManyToManyField called\nsites to filter on. If you use a field named something other than\nsite or sites to identify which\nSite objects your object is\nrelated to, then you need to explicitly pass the custom field name as\na parameter to\nCurrentSiteManager on your\nmodel. The following model, which has a field called publish_on,\ndemonstrates this:
from django.contrib.sites.models import Site\nfrom django.contrib.sites.managers import CurrentSiteManager\nfrom django.db import models\n\nclass Photo(models.Model):\n photo = models.FileField(upload_to='photos')\n photographer_name = models.CharField(max_length=100)\n pub_date = models.DateField()\n publish_on = models.ForeignKey(Site, on_delete=models.CASCADE)\n objects = models.Manager()\n on_site = CurrentSiteManager('publish_on')\nIf you attempt to use CurrentSiteManager\nand pass a field name that doesn't exist, Django will raise a ValueError.
Finally, note that you'll probably want to keep a normal\n(non-site-specific) Manager on your model, even if you use\nCurrentSiteManager. As\nexplained in the manager documentation, if\nyou define a manager manually, then Django won't create the automatic\nobjects = models.Manager() manager for you. Also note that certain\nparts of Django -- namely, the Django admin site and generic views --\nuse whichever manager is defined first in the model, so if you want\nyour admin site to have access to all objects (not just site-specific\nones), put objects = models.Manager() in your model, before you\ndefine CurrentSiteManager.
If you often use this pattern:
\nfrom django.contrib.sites.models import Site\n\ndef my_view(request):\n site = Site.objects.get_current()\n ...\nTo avoid repetitions, add\ndjango.contrib.sites.middleware.CurrentSiteMiddleware to\nMIDDLEWARE. The middleware sets the site attribute on every\nrequest 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\nencouraged, because Django takes advantage of it in a few places. Even if your\nDjango installation is powering only a single site, you should take the two\nseconds to create the site object with your domain and name, and point\nto its ID in your SITE_ID setting.
Here's how Django uses the sites framework:
\nIn the redirects framework, each\nredirect object is associated with a particular site. When Django searches\nfor a redirect, it takes into account the current site.
In the flatpages framework, each\nflatpage is associated with a particular site. When a flatpage is created,\nyou specify its Site, and the\nFlatpageFallbackMiddleware\nchecks the current site in retrieving flatpages to display.
In the syndication framework, the\ntemplates for title and description automatically have access to a\nvariable {{ site }}, which is the\nSite object representing the current\nsite. Also, the hook for providing item URLs will use the domain from\nthe current Site object if you don't\nspecify a fully-qualified domain.
In the authentication framework,\ndjango.contrib.auth.views.LoginView passes the current\nSite name to the template as\n{{ site_name }}.
The shortcut view (django.contrib.contenttypes.views.shortcut)\nuses the domain of the current\nSite object when calculating\nan object's URL.
In the admin framework, the "view on site" link uses the current\nSite to work out the domain for the\nsite that it will redirect to.
RequestSite objectsLink to this heading#Some django.contrib applications take advantage of\nthe sites framework but are architected in a way that doesn't require the\nsites framework to be installed in your database. (Some people don't want to,\nor just aren't able to install the extra database table that the sites\nframework requires.) For those cases, the framework provides a\ndjango.contrib.sites.requests.RequestSite class, which can be used as\na fallback when the database-backed sites framework is not available.
A class that shares the primary interface of\nSite (i.e., it has\ndomain and name attributes) but gets its data from a Django\nHttpRequest object rather than from a database.
Sets the name and domain attributes to the value of\nget_host().
A RequestSite object has a similar\ninterface to a normal Site object,\nexcept its __init__()\nmethod takes an HttpRequest object. It's able to deduce\nthe domain and name by looking at the request's domain. It has\nsave() and delete() methods to match the interface of\nSite, but the methods raise\nNotImplementedError.
get_current_site shortcutLink to this heading#Finally, to avoid repetitive fallback code, the framework provides a\ndjango.contrib.sites.shortcuts.get_current_site() function.
A function that checks if django.contrib.sites is installed and\nreturns either the current Site\nobject or a RequestSite object\nbased on the request. It looks up the current site based on\nrequest.get_host() if the\nSITE_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\nexplicitly specified, e.g. example.com:80. In such cases, if the\nlookup fails because the host does not match a record in the database,\nthe port is stripped and the lookup is retried with the domain part\nonly. This does not apply to\nRequestSite which will always\nuse the unmodified host.