Display tick and cross icons for a property in the Django administration console
Question:
In the Django admin if the field is a BooleanField or NullBooleanField, Django will display a pretty “on” or “off” icon instead of True or False.
Now, I don’t really have a BooleanField in my model by I do have a property fior which I’d like to display the icons but when I try doing so, Django screams that 'SomeAdmin.list_filter[0]' refers to 'is_activated' which does not refer to a Field.
Is it possible to display those nice little icons for this field without hacking Django too much.
Thanks
Answers:
You don’t want to use list_filter
. The property you’re looking for is list_display
. The documentation offers an example of how you can create a column that behaves like a boolean in the display. In short, you do something like this:
-
Create a method in the class:
def is_activated(self)
if self.bar == 'something':
return True
return False
-
add the .boolean
method attribute directly below the is_activated
method:
is_activated.boolean = True
-
Add the method as a field in list_display
:
class MyAdmin(ModelAdmin):
list_display = [‘name’, ‘is_activated’]
-
You’ll notice the column name is probably now “Is Activated” or something like that. If you want the column heading to change, you use the short_description
method attribute:
is_activated.short_description = "Activated"
The correct way to do this now in Django 3.0+ is with @admin.display
.
@admin.display(
boolean=True,
ordering='-publish_date',
description='Is Published?',
)
def is_published(self, obj):
return obj.publish_date is not None
In the Django admin if the field is a BooleanField or NullBooleanField, Django will display a pretty “on” or “off” icon instead of True or False.
Now, I don’t really have a BooleanField in my model by I do have a property fior which I’d like to display the icons but when I try doing so, Django screams that 'SomeAdmin.list_filter[0]' refers to 'is_activated' which does not refer to a Field.
Is it possible to display those nice little icons for this field without hacking Django too much.
Thanks
You don’t want to use list_filter
. The property you’re looking for is list_display
. The documentation offers an example of how you can create a column that behaves like a boolean in the display. In short, you do something like this:
-
Create a method in the class:
def is_activated(self) if self.bar == 'something': return True return False
-
add the
.boolean
method attribute directly below theis_activated
method:is_activated.boolean = True
-
Add the method as a field in
list_display
:class MyAdmin(ModelAdmin):
list_display = [‘name’, ‘is_activated’] -
You’ll notice the column name is probably now “Is Activated” or something like that. If you want the column heading to change, you use the
short_description
method attribute:is_activated.short_description = "Activated"
The correct way to do this now in Django 3.0+ is with @admin.display
.
@admin.display( boolean=True, ordering='-publish_date', description='Is Published?', ) def is_published(self, obj): return obj.publish_date is not None