Creating Custom Filters for list_filter in Django Admin
I would like to make custom filters for django admin instead of the normal 'is_staff' and 'is_superuser'. I have read this list_filter in Django docs. Custom Filters work in this way:
from datetime import date from django.utils.translation import ugettext_lazy as _ from django.contrib.admin import SimpleListFilter class DecadeBornListFilter(SimpleListFilter): # Human-readable title which will be displayed in the # right admin sidebar just above the filter options. title = _('decade born') # Parameter for the filter that will be used in the URL query. parameter_name = 'decade' def lookups(self, request, model_admin): """ Returns a list of tuples. The first element in each tuple is the coded value for the option that will appear in the URL query. The second element is the human-readable name for the option that will appear in the right sidebar. """ return ( ('80s', _('in the eighties')), ('90s', _('in the nineties')), ) def queryset(self, request, queryset): """ Returns the filtered queryset based on the value provided in the query string and retrievable via `self.value()`. """ # Compare the requested value (either '80s' or '90s') # to decide how to filter the queryset. if self.value() == '80s': return queryset.filter(birthday__gte=date(1980, 1, 1), birthday__lte=date(1989, 12, 31)) if self.value() == '90s': return queryset.filter(birthday__gte=date(1990, 1, 1), birthday__lte=date(1999, 12, 31)) class PersonAdmin(ModelAdmin): list_filter = (DecadeBornListFilter,)
But i have already made custom functions for list_display like this:
def Student_Country(self, obj): return '%s' % obj.country Student_Country.short_description = 'Student-Country'
Is it possible i could use the custom functions for list_display in list_filter instead of writing a new custom function for list_filter? Any suggestions or improvements are welcome.. Need some guidance on this... Thanks...
You can indeed add custom filters to admin filters by extending SimpleListFilter. For instance, if you want to add a continent filter for 'Africa' to the country admin filter used above, you can do the following:
from django.contrib.admin import SimpleListFilter class CountryFilter(SimpleListFilter): title = 'country' # or use _('country') for translated title parameter_name = 'country' def lookups(self, request, model_admin): countries = set([c.country for c in model_admin.model.objects.all()]) return [(c.id, c.name) for c in countries] + [ ('AFRICA', 'AFRICA - ALL')] def queryset(self, request, queryset): if self.value() == 'AFRICA': return queryset.filter(country__continent='Africa') if self.value(): return queryset.filter(country__id__exact=self.value()) class CityAdmin(ModelAdmin): list_filter = (CountryFilter,)
Your list_display ,method returns a string, but if I understand correctly, what you want to do is add a filter which allows selection of countries of students, correct?
For this simple relation filter, and in fact for the "Student-Country" list display column as well, you don't need to create a custom filter class, nor a custom list display method; this would suffice:
class MyAdmin(admin.ModelAdmin): list_display = ('country', ) list_filter = ('country', )
The way django does list_filter, as explained in the docs, is first by automatically matching fields you provide to pre-built filter classes; these filters include CharField and ForeignKey.
list_display similarly automates the population of the changelist column using the field passed by retrieving the related objects and returning the unicode value of these (same as in the method you provided above).
ModelAdmin.list_filter Set list_filter to activate filters in the right sidebar of the change list page of the admin list_filter should be a list or tuple of elements