sites.managers.CurrentSiteManager

class 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.

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, 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.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, 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.