By default Django admin site shows all records of a related model/table for viewing. How can I show only the records that meet certain criteria?
Django admin allows access to users marked as is_staff=True . To disable a user from being able to access the admin, you should set is_staff=False . This holds true even if the user is a superuser. is_superuser=True .
For Django 1.4-1.7, list_filter allows you to use a subclass of SimpleListFilter . It should be possible to create a simple list filter that lists the values you want.
Overview. The Django admin application can use your models to automatically build a site area that you can use to create, view, update, and delete records. This can save you a lot of time during development, making it very easy to test your models and get a feel for whether you have the right data.
In your admin definition, you can define a queryset()
method that returns the queryset for that model's admin. eg:
class MyModelAdmin(admin.ModelAdmin):
def queryset(self, request):
qs = super(MyModelAdmin, self).queryset(request)
return qs.filter(user=request.user)
Then only objects with user=request.user
will be visible in the admin.
I know this has an "accepted answer", but I just wanted to throw this out there since I came across this answer while pursuing something else and realized I had an alternative solution that I found and use often that gives me more granular level control than the accepted answer.
class TestAdmin(admin.ModelAdmin):
def formfield_for_foreignkey(self, db_field, request, **kwargs):
if db_field.name == "FIELD":
kwargs["queryset"] = TestModel.objects.filter(test=False)
return super(TestAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs)
def formfield_for_manytomany(self, db_field, request, **kwargs):
if db_field.name == "FIELDS":
kwargs["queryset"] = TestModel.objects.filter(test=False)
return super(TestAdmin, self).formfield_for_manytomany(db_field, request, **kwargs)
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