Programming

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====== Sites framework ======

  * https://docs.djangoproject.com/en/3.0/ref/contrib/sites

Based on the ''Site'' class which its important attribute is ''domain'' that specifies the domain name of the site.

===== Usage =====

  - 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.


===== Basic example =====

<code python>
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)
    # or: site = models.ForeignKey(Site, on_delete=models.CASCADE)

# ...

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")
    # ...
</code>

===== Current site =====
<code python>
from django.contrib.sites.shortcuts import get_current_site
# ...
current_site = get_current_site(request)
</code>


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.

<code python>
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
</code>

==== The middleware ====

To add the ''site'' attribute to the ''request'' for a view just add ''django.contrib.sites.middleware.CurrentSiteMiddleware'' to ''MIDDLEWARE''.
==== The full url ====
<code python>
>>> 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/'
</code>


===== Beheaving depending on the site =====
<code python>
from django.conf import settings

def my_view(request):
    if settings.SITE_ID == 3:
        # Do something.
        pass
    else:
        # Do something else.
        pass
</code>
or...
<code python>
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
</code> 

If you wanted to choose different templates refer to [[https://docs.djangoproject.com/en/3.0/ref/contrib/sites/#getting-the-current-domain-for-display|this link]].

==== Querying depending on the site ====
<code python>
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()
</code>

Having this we can:
  * Query all Photo objects in the database with ''Photo.objects.all()''. 
  * Query only the Photo objects associated with the current site (according to the ''SITE_ID''): ''Photo.on_site.all()''
Programming

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