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This work started life as Brian Beck's "django-batchadmin." It was rewritten for inclusion in Django by Alex Gaynor, Jannis Leidel (jezdez), and Martin Mahner (bartTC). Thanks, guys! git-svn-id: http://code.djangoproject.com/svn/django/trunk@10121 bcc190cf-cafb-0310-a4f2-bffc1f526a37
240 lines
9.0 KiB
Plaintext
240 lines
9.0 KiB
Plaintext
.. _ref-contrib-admin-actions:
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=============
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Admin actions
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=============
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.. versionadded:: 1.1
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.. currentmodule:: django.contrib.admin
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The basic workflow of Django's admin is, in a nutshell, "select an object,
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then change it." This works well for a majority of use cases. However, if you
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need to make the same change to many objects at once, this workflow can be
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quite tedious.
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In these cases, Django's admin lets you write and register "actions" -- simple
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functions that get called with a list of objects selected on the change list
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page.
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If you look at any change list in the admin, you'll see this feature in
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action; Django ships with a "delete selected objects" action available to all
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models. For example, here's the user module from Django's built-in
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:mod:`django.contrib.auth` app:
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.. image:: _images/user_actions.png
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Read on to find out how to add your own actions to this list.
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Writing actions
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===============
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The easiest way to explain actions is by example, so let's dive in.
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A common use case for admin actions is the bulk updating of a model. Imagine a simple
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news application with an ``Article`` model::
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from django.db import models
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STATUS_CHOICES = (
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('d', 'Draft'),
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('p', 'Published'),
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('w', 'Withdrawn'),
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)
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class Article(models.Model):
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title = models.CharField(max_length=100)
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body = models.TextField()
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status = models.CharField(max_length=1, choices=STATUS_CHOICES)
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def __unicode__(self):
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return self.title
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A common task we might perform with a model like this is to update an
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article's status from "draft" to "published". We could easily do this in the
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admin one article at a time, but if we wanted to bulk-publish a group of
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articles, it'd be tedious. So, let's write an action that lets us change an
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article's status to "published."
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Writing action functions
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------------------------
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First, we'll need to write a function that gets called when the action is
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trigged from the admin. Action functions are just regular functions that take
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two arguments: an :class:`~django.http.HttpRequest` representing the current
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request, and a :class:`~django.db.models.QuerySet` containing the set of
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objects selected by the user. Our publish-these-articles function won't need
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the request object, but we will use the queryset::
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def make_published(request, queryset):
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queryset.update(status='p')
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.. note::
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For the best performance, we're using the queryset's :ref:`update method
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<topics-db-queries-update>`. Other types of actions might need to deal
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with each object individually; in these cases we'd just iterate over the
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queryset::
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for obj in queryset:
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do_something_with(obj)
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That's actually all there is to writing an action! However, we'll take one
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more optional-but-useful step and give the action a "nice" title in the admin.
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By default, this action would appear in the action list as "Make published" --
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the function name, with underscores replaced by spaces. That's fine, but we
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can provide a better, more human-friendly name by giving the
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``make_published`` function a ``short_description`` attribute::
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def make_published(request, queryset):
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queryset.update(status='p')
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make_published.short_description = "Mark selected stories as published"
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.. note::
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This might look familiar; the admin's ``list_display`` option uses the
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same technique to provide human-readable descriptions for callback
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functions registered there, too.
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Adding actions to the :class:`ModelAdmin`
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-----------------------------------------
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Next, we'll need to inform our :class:`ModelAdmin` of the action. This works
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just like any other configuration option. So, the complete ``admin.py`` with
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the action and its registration would look like::
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from django.contrib import admin
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from myapp.models import Article
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def make_published(request, queryset):
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queryset.update(status='p')
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make_published.short_description = "Mark selected stories as published"
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class ArticleAdmin(admin.ModelAdmin):
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list_display = ['title', 'status']
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ordering = ['title']
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actions = [make_published]
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admin.site.register(Article, ArticleAdmin)
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That code will give us an admin change list that looks something like this:
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.. image:: _images/article_actions.png
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That's really all there is to it! If you're itching to write your own actions,
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you now know enough to get started. The rest of this document just covers more
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advanced techniques.
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Advanced action techniques
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==========================
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There's a couple of extra options and possibilities you can exploit for more
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advanced options.
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Actions as :class:`ModelAdmin` methods
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--------------------------------------
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The example above shows the ``make_published`` action defined as a simple
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function. That's perfectly fine, but it's not perfect from a code design point
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of view: since the action is tightly coupled to the ``Article`` object, it
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makes sense to hook the action to the ``ArticleAdmin`` object itself.
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That's easy enough to do::
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class ArticleAdmin(admin.ModelAdmin):
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...
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actions = ['make_published']
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def make_published(self, request, queryset):
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queryset.update(status='p')
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make_published.short_description = "Mark selected stories as published"
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Notice first that we've moved ``make_published`` into a method (remembering to
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add the ``self`` argument!), and second that we've now put the string
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``'make_published'`` in ``actions`` instead of a direct function reference.
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This tells the :class:`ModelAdmin` to look up the action as a method.
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Defining actions as methods is especially nice because it gives the action
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access to the :class:`ModelAdmin` itself, allowing the action to call any of
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the methods provided by the admin.
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For example, we can use ``self`` to flash a message to the user informing her
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that the action was successful::
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class ArticleAdmin(admin.ModelAdmin):
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...
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def make_published(self, request, queryset):
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rows_updated = queryset.update(status='p')
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if rows_updated == 1:
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message_bit = "1 story was"
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else:
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message_bit = "%s stories were" % rows_updated
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self.message_user(request, "%s successfully marked as published." % message_bit)
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This make the action match what the admin itself does after successfully
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performing an action:
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.. image:: _images/article_actions_message.png
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Actions that provide intermediate pages
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---------------------------------------
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By default, after an action is performed the user is simply redirected back
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the the original change list page. However, some actions, especially more
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complex ones, will need to return intermediate pages. For example, the
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built-in delete action asks for confirmation before deleting the selected
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objects.
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To provide an intermediary page, simply return an
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:class:`~django.http.HttpResponse` (or subclass) from your action. For
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example, you might write a simple export function that uses Django's
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:ref:`serialization functions <topics-serialization>` to dump some selected
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objects as JSON::
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from django.http import HttpResponse
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from django.core import serializers
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def export_as_json(request, queryset):
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response = HttpResponse(mimetype="text/javascript")
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serialize.serialize(queryset, stream=response)
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return response
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Generally, something like the above isn't considered a great idea. Most of the
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time, the best practice will be to return an
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:class:`~django.http.HttpResponseRedirect` and redirect the user to a view
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you've written, passing the list of selected objects in the GET query string.
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This allows you to provide complex interaction logic on the intermediary
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pages. For example, if you wanted to provide a more complete export function,
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you'd want to let the user choose a format, and possibly a list of fields to
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include in the export. The best thing to do would be to write a small action that simply redirects
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to your custom export view::
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from django.contrib import admin
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from django.contrib.contenttypes.models import ContentType
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from django.http import HttpResponseRedirect
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def export_selected_objects(request, queryset):
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selected = request.POST.getlist(admin.ACTION_CHECKBOX_NAME)
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ct = ContentType.objects.get_for_model(queryset.model)
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return HttpResponseRedirect("/export/?ct=%s&ids=%s" % (ct.pk, ",".join(selected)))
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As you can see, the action is the simple part; all the complex logic would
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belong in your export view. This would need to deal with objects of any type,
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hence the business with the ``ContentType``.
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Writing this view is left as an exercise to the reader.
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Making actions available globally
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---------------------------------
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Some actions are best if they're made available to *any* object in the admin
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-- the export action defined above would be a good candidate. You can make an
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action globally available using :meth:`AdminSite.add_action()`::
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from django.contrib import admin
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admin.site.add_action(export_selected_objects)
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