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
The “delete selected objects” action uses
QuerySet.delete() for efficiency reasons, which
has an important caveat: your model’s
delete() method will not be
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
from django.db import models STATUS_CHOICES = [ ('d', 'Draft'), ('p', 'Published'), ('w', 'Withdrawn'), ] class Article(models.Model): title = models.CharField(max_length=100) body = models.TextField() status = models.CharField(max_length=1,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:
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')
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
def make_published(modeladmin, request, queryset): queryset.update(status='p') make_published.short_description = "Mark selected stories as published"
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.
Next, we’ll need to inform our
ModelAdmin of the action. This works
just like any other configuration option. So, the complete
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.
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
actions instead of a direct function reference. This
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.contenttypes.models import ContentType from django.http import HttpResponseRedirect def export_selected_objects(modeladmin, request, queryset): selected = queryset.values_list('pk', flat=True) ct = ContentType.objects.get_for_model(queryset.model) return HttpResponseRedirect('/export/?ct=%s&ids=%s' % ( ct.pk, ','.join(str(pk) for pk in 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
Writing this view is left as an exercise to the reader.
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
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
Sometimes you need to disable certain actions – especially those registered site-wide – for particular objects. There’s a few ways you can disable actions:
For example, you can use this method to remove the built-in “delete selected objects” action:
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
# 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'] ...
class MyModelAdmin(admin.ModelAdmin): actions = None
Finally, you can conditionally enable or disable actions on a per-request
(and hence per-user basis) by overriding
This returns a dictionary of actions allowed. The keys are action names, and
the values are
(function, name, short_description) tuples.
For example, if you only want users whose names begin with ‘J’ to be able to delete objects in bulk:
class MyModelAdmin(admin.ModelAdmin): ... def get_actions(self, request): actions = super().get_actions(request) if request.user.username.upper() != 'J': if 'delete_selected' in actions: del actions['delete_selected'] return actions
Actions may limit their availability to users with specific permissions by
allowed_permissions attribute on the action function:
def make_published(modeladmin, request, queryset): queryset.update(status='p') make_published.allowed_permissions = ('change',)
make_published() action will only be available to users that pass the
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
You can specify any other value as long as you implement a corresponding
has_<value>_permission(self, request) method on the
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))