The basic workflow of Django’s admin is, in a nutshell, “select an object, then change it.” This works well for a majority of use cases. However, if you need to make the same change to many objects at once, this workflow can be quite tedious.
In these cases, Django’s admin lets you write and register “actions” – simple functions that get called with a list of objects selected on the change list page.
If you look at any change list in the admin, you’ll see this feature in
action; Django ships with a “delete selected objects” action available to all
models. For example, here’s the user module from Django’s built-in
django.contrib.auth
app:
Warning
The “delete selected objects” action uses QuerySet.delete()
for efficiency reasons, which
has an important caveat: your model’s delete()
method will not be
called.
If you wish to override this behavior, you can override
ModelAdmin.delete_queryset()
or write a custom action which does
deletion in your preferred manner – for example, by calling
Model.delete()
for each of the selected items.
For more background on bulk deletion, see the documentation on object deletion.
Read on to find out how to add your own actions to this list.
The easiest way to explain actions is by example, so let’s dive in.
A common use case for admin actions is the bulk updating of a model. Imagine a
simple news application with an Article
model:
from django.db import models
STATUS_CHOICES = (
('d', 'Draft'),
('p', 'Published'),
('w', 'Withdrawn'),
)
class Article(models.Model):
title = models.CharField(max_length=100)
body = models.TextField()
status = models.CharField(max_length=1, choices=STATUS_CHOICES)
def __str__(self):
return self.title
A common task we might perform with a model like this is to update an article’s status from “draft” to “published”. We could easily do this in the admin one article at a time, but if we wanted to bulk-publish a group of articles, it’d be tedious. So, let’s write an action that lets us change an article’s status to “published.”
First, we’ll need to write a function that gets called when the action is triggered from the admin. Action functions are just regular functions that take three arguments:
ModelAdmin
HttpRequest
representing the current request,QuerySet
containing the set of
objects selected by the user.Our publish-these-articles function won’t need the ModelAdmin
or the
request object, but we will use the queryset:
def make_published(modeladmin, request, queryset):
queryset.update(status='p')
Note
For the best performance, we’re using the queryset’s update method. Other types of actions might need to deal with each object individually; in these cases we’d just iterate over the queryset:
for obj in queryset:
do_something_with(obj)
That’s actually all there is to writing an action! However, we’ll take one
more optional-but-useful step and give the action a “nice” title in the admin.
By default, this action would appear in the action list as “Make published” –
the function name, with underscores replaced by spaces. That’s fine, but we
can provide a better, more human-friendly name by giving the
make_published
function a short_description
attribute:
def make_published(modeladmin, request, queryset):
queryset.update(status='p')
make_published.short_description = "Mark selected stories as published"
Note
This might look familiar; the admin’s list_display
option uses the
same technique to provide human-readable descriptions for callback
functions registered there, too.
ModelAdmin
¶Next, we’ll need to inform our ModelAdmin
of the action. This works
just like any other configuration option. So, the complete admin.py
with
the action and its registration would look like:
from django.contrib import admin
from myapp.models import Article
def make_published(modeladmin, request, queryset):
queryset.update(status='p')
make_published.short_description = "Mark selected stories as published"
class ArticleAdmin(admin.ModelAdmin):
list_display = ['title', 'status']
ordering = ['title']
actions = [make_published]
admin.site.register(Article, ArticleAdmin)
That code will give us an admin change list that looks something like this:
That’s really all there is to it! If you’re itching to write your own actions, you now know enough to get started. The rest of this document just covers more advanced techniques.
If there are foreseeable error conditions that may occur while running your
action, you should gracefully inform the user of the problem. This means
handling exceptions and using
django.contrib.admin.ModelAdmin.message_user()
to display a user friendly
description of the problem in the response.
There’s a couple of extra options and possibilities you can exploit for more advanced options.
ModelAdmin
methods¶The example above shows the make_published
action defined as a simple
function. That’s perfectly fine, but it’s not perfect from a code design point
of view: since the action is tightly coupled to the Article
object, it
makes sense to hook the action to the ArticleAdmin
object itself.
That’s easy enough to do:
class ArticleAdmin(admin.ModelAdmin):
...
actions = ['make_published']
def make_published(self, request, queryset):
queryset.update(status='p')
make_published.short_description = "Mark selected stories as published"
Notice first that we’ve moved make_published
into a method and renamed the
modeladmin
parameter to self
, and second that we’ve now put the string
'make_published'
in actions
instead of a direct function reference. This
tells the ModelAdmin
to look up the action as a method.
Defining actions as methods gives the action more straightforward, idiomatic
access to the ModelAdmin
itself, allowing the action to call any of the
methods provided by the admin.
For example, we can use self
to flash a message to the user informing her
that the action was successful:
class ArticleAdmin(admin.ModelAdmin):
...
def make_published(self, request, queryset):
rows_updated = queryset.update(status='p')
if rows_updated == 1:
message_bit = "1 story was"
else:
message_bit = "%s stories were" % rows_updated
self.message_user(request, "%s successfully marked as published." % message_bit)
This make the action match what the admin itself does after successfully performing an action:
By default, after an action is performed the user is simply redirected back to the original change list page. However, some actions, especially more complex ones, will need to return intermediate pages. For example, the built-in delete action asks for confirmation before deleting the selected objects.
To provide an intermediary page, simply return an
HttpResponse
(or subclass) from your action. For
example, you might write a simple export function that uses Django’s
serialization functions to dump some selected
objects as JSON:
from django.core import serializers
from django.http import HttpResponse
def export_as_json(modeladmin, request, queryset):
response = HttpResponse(content_type="application/json")
serializers.serialize("json", queryset, stream=response)
return response
Generally, something like the above isn’t considered a great idea. Most of the
time, the best practice will be to return an
HttpResponseRedirect
and redirect the user to a view
you’ve written, passing the list of selected objects in the GET query string.
This allows you to provide complex interaction logic on the intermediary
pages. For example, if you wanted to provide a more complete export function,
you’d want to let the user choose a format, and possibly a list of fields to
include in the export. The best thing to do would be to write a small action
that simply redirects to your custom export view:
from django.contrib import admin
from django.contrib.contenttypes.models import ContentType
from django.http import HttpResponseRedirect
def export_selected_objects(modeladmin, request, queryset):
selected = request.POST.getlist(admin.ACTION_CHECKBOX_NAME)
ct = ContentType.objects.get_for_model(queryset.model)
return HttpResponseRedirect("/export/?ct=%s&ids=%s" % (ct.pk, ",".join(selected)))
As you can see, the action is the simple part; all the complex logic would
belong in your export view. This would need to deal with objects of any type,
hence the business with the ContentType
.
Writing this view is left as an exercise to the reader.
AdminSite.
add_action
(action, name=None)[source]¶Some actions are best if they’re made available to any object in the admin
site – the export action defined above would be a good candidate. You can
make an action globally available using AdminSite.add_action()
. For
example:
from django.contrib import admin
admin.site.add_action(export_selected_objects)
This makes the export_selected_objects
action globally available as an
action named “export_selected_objects”. You can explicitly give the action
a name – good if you later want to programmatically remove the action – by passing a second argument to
AdminSite.add_action()
:
admin.site.add_action(export_selected_objects, 'export_selected')
Sometimes you need to disable certain actions – especially those registered site-wide – for particular objects. There’s a few ways you can disable actions:
AdminSite.
disable_action
(name)[source]¶If you need to disable a site-wide action you can
call AdminSite.disable_action()
.
For example, you can use this method to remove the built-in “delete selected objects” action:
admin.site.disable_action('delete_selected')
Once you’ve done the above, that action will no longer be available site-wide.
If, however, you need to re-enable a globally-disabled action for one
particular model, simply list it explicitly in your ModelAdmin.actions
list:
# Globally disable delete selected
admin.site.disable_action('delete_selected')
# This ModelAdmin will not have delete_selected available
class SomeModelAdmin(admin.ModelAdmin):
actions = ['some_other_action']
...
# This one will
class AnotherModelAdmin(admin.ModelAdmin):
actions = ['delete_selected', 'a_third_action']
...
ModelAdmin
¶If you want no bulk actions available for a given ModelAdmin
, simply
set ModelAdmin.actions
to None
:
class MyModelAdmin(admin.ModelAdmin):
actions = None
This tells the ModelAdmin
to not display or allow any actions,
including any site-wide actions.
ModelAdmin.
get_actions
(request)[source]¶Finally, you can conditionally enable or disable actions on a per-request
(and hence per-user basis) by overriding ModelAdmin.get_actions()
.
This returns a dictionary of actions allowed. The keys are action names, and
the values are (function, name, short_description)
tuples.
Most of the time you’ll use this method to conditionally remove actions from the list gathered by the superclass. For example, if I only wanted users whose names begin with ‘J’ to be able to delete objects in bulk, I could do the following:
class MyModelAdmin(admin.ModelAdmin):
...
def get_actions(self, request):
actions = super().get_actions(request)
if request.user.username[0].upper() != 'J':
if 'delete_selected' in actions:
del actions['delete_selected']
return actions
Actions may limit their availability to users with specific permissions by
setting an allowed_permissions
attribute on the action function:
def make_published(modeladmin, request, queryset):
queryset.update(status='p')
make_published.allowed_permissions = ('change',)
The make_published()
action will only be available to users that pass the
ModelAdmin.has_change_permission()
check.
If allowed_permissions
has more than one permission, the action will be
available as long as the user passes at least one of the checks.
Available values for allowed_permissions
and the corresponding method
checks are:
'add'
: ModelAdmin.has_add_permission()
'change'
: ModelAdmin.has_change_permission()
'delete'
: ModelAdmin.has_delete_permission()
'view'
: ModelAdmin.has_view_permission()
You can specify any other value as long as you implement a corresponding
has_<value>_permission(self, request)
method on the ModelAdmin
.
For example:
from django.contrib import admin
from django.contrib.auth import get_permission_codename
class ArticleAdmin(admin.ModelAdmin):
actions = ['make_published']
def make_published(self, request, queryset):
queryset.update(status='p')
make_published.allowed_permissions = ('publish',)
def has_publish_permission(self, request):
"""Does the user have the publish permission?"""
opts = self.opts
codename = get_permission_codename('publish', opts)
return request.user.has_perm('%s.%s' % (opts.app_label, codename))
Jul 24, 2018