How do I override an admin template (e.g. admin/index.html) while at the same time extending it (see https://docs.djangoproject.com/en/dev/ref/contrib/admin/#overriding-vs-replacing-an-admin-template)?

First – I know that this question has been asked and answered before (see Django: Overriding AND extending an app template) but as the answer says it isn’t directly applicable if you’re using the app_directories template loader (which is most of the time).

My current workaround is to make copies and extend from them instead of extending directly from the admin templates. This works great but it’s really confusing and adds extra work when the admin templates change.

It could think of some custom extend-tag for the templates but I don’t want to reinvent the wheel if there already exists a solution.

On a side note: Does anybody know if this problem will be addressed by Django itself?

Update:

Read the Docs for your version of Django. e.g.

https://docs.djangoproject.com/en/1.11/ref/contrib/admin/#admin-overriding-templates https://docs.djangoproject.com/en/2.0/ref/contrib/admin/#admin-overriding-templates https://docs.djangoproject.com/en/3.0/ref/contrib/admin/#admin-overriding-templates

Original answer from 2011:

I had the same issue about a year and a half ago and I found a nice template loader on djangosnippets.org that makes this easy. It allows you to extend a template in a specific app, giving you the ability to create your own admin/index.html that extends the admin/index.html template from the admin app. Like this:

{% extends "admin:admin/index.html" %}

{% block sidebar %}
    {{block.super}}
    <div>
        <h1>Extra links</h1>
        <a href="/admin/extra/">My extra link</a>
    </div>
{% endblock %}

I’ve given a full example on how to use this template loader in a blog post on my website.

As for Django 1.8 being the current release, there is no need to symlink, copy the admin/templates to your project folder, or install middlewares as suggested by the answers above. Here is what to do:

1. create the following tree structure(recommended by the official documentation)

   your_project
        |-- your_project/
        |-- myapp/
        |-- templates/
             |-- admin/
                 |-- myapp/
                     |-- change_form.html  <- do not misspell this
   

Note: The location of this file is not important. You can put it inside your app and it will still work. As long as its location can be discovered by django. What’s more important is the name of the HTML file has to be the same as the original HTML file name provided by django.

2. Add this template path to your settings.py:

   TEMPLATES = [
       {
           'BACKEND': 'django.template.backends.django.DjangoTemplates',
           'DIRS': [os.path.join(BASE_DIR, 'templates')], # <- add this line
           'APP_DIRS': True,
           'OPTIONS': {
               'context_processors': [
                   'django.template.context_processors.debug',
                   'django.template.context_processors.request',
                   'django.contrib.auth.context_processors.auth',
                   'django.contrib.messages.context_processors.messages',
               ],
           },
       },
   ]
   

3. Identify the name and block you want to override. This is done by looking into django’s admin/templates directory. I am using virtualenv, so for me, the path is here:

   ~/.virtualenvs/edge/lib/python2.7/site-packages/django/contrib/admin/templates/admin
   

In this example, I want to modify the add new user form. The template responsiblve for this view is change_form.html. Open up the change_form.html and find the {% block %} that you want to extend.

4. In your change_form.html, write somethings like this:

   {% extends "admin/change_form.html" %}
   {% block field_sets %}
        {# your modification here #}
   {% endblock %}
   

5. Load up your page and you should see the changes

if you need to overwrite the admin/index.html, you can set the index_template parameter of the AdminSite.

e.g.

# urls.py
...
from django.contrib import admin

admin.site.index_template = 'admin/my_custom_index.html'
admin.autodiscover()

and place your template in <appname>/templates/admin/my_custom_index.html

With django 1.5 (at least) you can define the template you want to use for a particular modeladmin

see https://docs.djangoproject.com/en/1.5/ref/contrib/admin/#custom-template-options

You can do something like

class Myadmin(admin.ModelAdmin):
    change_form_template = 'change_form.htm'

With change_form.html being a simple html template extending admin/change_form.html (or not if you want to do it from scratch)

Chengs’s answer is correct, howewer according to the admin docs not every admin template can be overwritten this way: https://docs.djangoproject.com/en/1.9/ref/contrib/admin/#overriding-admin-templates

Templates which may be overridden per app or model

Not every template in contrib/admin/templates/admin may be overridden per app or per model. The following can:

app_index.html
change_form.html
change_list.html
delete_confirmation.html
object_history.html

For those templates that cannot be overridden in this way, you may still override them for your entire project. Just place the new version in your templates/admin directory. This is particularly useful to create custom 404 and 500 pages

I had to overwrite the login.html of the admin and therefore had to put the overwritten template in this folder structure:

your_project
 |-- your_project/
 |-- myapp/
 |-- templates/
      |-- admin/
          |-- login.html  <- do not misspell this

(without the myapp subfolder in the admin) I do not have enough repution for commenting on Cheng’s post this is why I had to write this as new answer.

The best way to do it is to put the Django admin templates inside your project. So your templates would be in templates/admin while the stock Django admin templates would be in say template/django_admin. Then, you can do something like the following:

templates/admin/change_form.html

{% extends 'django_admin/change_form.html' %}

Your stuff here

If you’re worried about keeping the stock templates up to date, you can include them with svn externals or similar.

I couldn’t find a single answer or a section in the official Django docs that had all the information I needed to override/extend the default admin templates, so I’m writing this answer as a complete guide, hoping that it would be helpful for others in the future.

Assuming the standard Django project structure:

mysite-container/         # project container directory
    manage.py
    mysite/               # project package
        __init__.py
        admin.py
        apps.py
        settings.py
        urls.py
        wsgi.py
    app1/
    app2/
    ...
    static/
    templates/

Here’s what you need to do:

1. In mysite/admin.py, create a sub-class of AdminSite:

   from django.contrib.admin import AdminSite
   
   
   class CustomAdminSite(AdminSite):
       # set values for `site_header`, `site_title`, `index_title` etc.
       site_header = 'Custom Admin Site'
       ...
   
       # extend / override admin views, such as `index()`
       def index(self, request, extra_context=None):
           extra_context = extra_context or {}
   
           # do whatever you want to do and save the values in `extra_context`
           extra_context['world'] = 'Earth'
   
           return super(CustomAdminSite, self).index(request, extra_context)
   
   
   custom_admin_site = CustomAdminSite()
   

   Make sure to import custom_admin_site in the admin.py of your apps and register your models on it to display them on your customized admin site (if you want to).

2. In mysite/apps.py, create a sub-class of AdminConfig and set default_site to admin.CustomAdminSite from the previous step:

   from django.contrib.admin.apps import AdminConfig
   
   
   class CustomAdminConfig(AdminConfig):
       default_site = 'admin.CustomAdminSite'
   

3. In mysite/settings.py, replace django.admin.site in INSTALLED_APPS with apps.CustomAdminConfig (your custom admin app config from the previous step).

4. In mysite/urls.py, replace admin.site.urls from the admin URL to custom_admin_site.urls

   from .admin import custom_admin_site
   
   
   urlpatterns = [
       ...
       path('admin/', custom_admin_site.urls),
       # for Django 1.x versions: url(r'^admin/', include(custom_admin_site.urls)),
       ...
   ]
   

5. Create the template you want to modify in your templates directory, maintaining the default Django admin templates directory structure as specified in the docs. For example, if you were modifying admin/index.html, create the file templates/admin/index.html.

   All of the existing templates can be modified this way, and their names and structures can be found in Django’s source code.

6. Now you can either override the template by writing it from scratch or extend it and then override/extend specific blocks.

   For example, if you wanted to keep everything as-is but wanted to override the content block (which on the index page lists the apps and their models that you registered), add the following to templates/admin/index.html:

   {% extends 'admin/index.html' %}
   
   {% block content %}
     <h1>
       Hello, {{ world }}!
     </h1>
   {% endblock %}
   

   To preserve the original contents of a block, add {{ block.super }} wherever you want the original contents to be displayed:

   {% extends 'admin/index.html' %}
   
   {% block content %}
     <h1>
       Hello, {{ world }}!
     </h1>
     {{ block.super }}
   {% endblock %}
   

   You can also add custom styles and scripts by modifying the extrastyle and extrahead blocks.

I agree with Chris Pratt. But I think it’s better to create the symlink to original Django folder where the admin templates place in:

ln -s /usr/local/lib/python2.7/dist-packages/django/contrib/admin/templates/admin/ templates/django_admin

and as you can see it depends on python version and the folder where the Django installed. So in future or on a production server you might need to change the path.

This site had a simple solution that worked with my Django 1.7 configuration.

FIRST: Make a symlink named admin_src in your project’s template/ directory to your installed Django templates. For me on Dreamhost using a virtualenv, my “source” Django admin templates were in:

~/virtualenvs/mydomain/lib/python2.7/site-packages/django/contrib/admin/templates/admin

SECOND: Create an admin directory in templates/

So my project’s template/ directory now looked like this:

/templates/
   admin
   admin_src -> [to django source]
   base.html
   index.html
   sitemap.xml
   etc...

THIRD: In your new template/admin/ directory create a base.html file with this content:

{% extends "admin_src/base.html" %}

{% block extrahead %}
<link rel='shortcut icon' href='{{ STATIC_URL }}img/favicon-admin.ico' />
{% endblock %}

FOURTH: Add your admin favicon-admin.ico into your static root img folder.

Done. Easy.

for app index add this line to somewhere common py file like url.py

admin.site.index_template = 'admin/custom_index.html'

for app module index : add this line to admin.py

admin.AdminSite.app_index_template = "servers/servers-home.html"

for change list : add this line to admin class:

change_list_template = "servers/servers_changelist.html"

for app module form template : add this line to your admin class

change_form_template = "servers/server_changeform.html"

etc. and find other in same admin’s module classes

You can use django-overextends, which provides circular template inheritance for Django.

It comes from the Mezzanine CMS, from where Stephen extracted it into a standalone Django extension.

More infos you find in “Overriding vs Extending Templates” (http:/mezzanine.jupo.org/docs/content-architecture.html#overriding-vs-extending-templates) inside the Mezzanine docs.

For deeper insides look at Stephens Blog “Circular Template Inheritance for Django” (http:/blog.jupo.org/2012/05/17/circular-template-inheritance-for-django).

And in Google Groups the discussion (https:/groups.google.com/forum/#!topic/mezzanine-users/sUydcf_IZkQ) which started the development of this feature.

Note:

I don’t have the reputation to add more than 2 links. But I think the links provide interesting background information. So I just left out a slash after “http(s):”. Maybe someone with better reputation can repair the links and remove this note.

How can I change the default filter choice from ‘ALL’? I have a field named as status which has three values: activate, pending and rejected. When I use list_filter in Django admin, the filter is by default set to ‘All’ but I want to set it to pending by default.

In order to achieve this and have a usable ‘All’ link in your sidebar (ie one that shows all rather than showing pending), you’d need to create a custom list filter, inheriting from django.contrib.admin.filters.SimpleListFilter and filtering on ‘pending’ by default. Something along these lines should work:

from datetime import date

from django.utils.translation import ugettext_lazy as _
from django.contrib.admin import SimpleListFilter

class StatusFilter(SimpleListFilter):
    title = _('Status')

    parameter_name = 'status'

    def lookups(self, request, model_admin):
        return (
            (None, _('Pending')),
            ('activate', _('Activate')),
            ('rejected', _('Rejected')),
            ('all', _('All')),
        )

    def choices(self, cl):
        for lookup, title in self.lookup_choices:
            yield {
                'selected': self.value() == lookup,
                'query_string': cl.get_query_string({
                    self.parameter_name: lookup,
                }, []),
                'display': title,
            }

    def queryset(self, request, queryset):
        if self.value() in ('activate', 'rejected'):
            return queryset.filter(status=self.value())    
        elif self.value() == None:
            return queryset.filter(status='pending')


class Admin(admin.ModelAdmin): 
    list_filter = [StatusFilter] 

EDIT: Requires Django 1.4 (thanks Simon)

class MyModelAdmin(admin.ModelAdmin):   

    def changelist_view(self, request, extra_context=None):

        if not request.GET.has_key('decommissioned__exact'):

            q = request.GET.copy()
            q['decommissioned__exact'] = 'N'
            request.GET = q
            request.META['QUERY_STRING'] = request.GET.urlencode()
        return super(MyModelAdmin,self).changelist_view(request, extra_context=extra_context)

Took ha22109’s answer above and modified to allow the selection of “All” by comparing HTTP_REFERER and PATH_INFO.

class MyModelAdmin(admin.ModelAdmin):

    def changelist_view(self, request, extra_context=None):

        test = request.META['HTTP_REFERER'].split(request.META['PATH_INFO'])

        if test[-1] and not test[-1].startswith('?'):
            if not request.GET.has_key('decommissioned__exact'):

                q = request.GET.copy()
                q['decommissioned__exact'] = 'N'
                request.GET = q
                request.META['QUERY_STRING'] = request.GET.urlencode()
        return super(MyModelAdmin,self).changelist_view(request, extra_context=extra_context)

I know this question is quite old now, but it’s still valid. I believe this is the most correct way of doing this. It’s essentially the same as Greg’s method, but formulated as an extendible class for easy re-use.

from django.contrib.admin import SimpleListFilter
from django.utils.encoding import force_text
from django.utils.translation import ugettext as _

class DefaultListFilter(SimpleListFilter):
    all_value = '_all'

    def default_value(self):
        raise NotImplementedError()

    def queryset(self, request, queryset):
        if self.parameter_name in request.GET and request.GET[self.parameter_name] == self.all_value:
            return queryset

        if self.parameter_name in request.GET:
            return queryset.filter(**{self.parameter_name:request.GET[self.parameter_name]})

        return queryset.filter(**{self.parameter_name:self.default_value()})

    def choices(self, cl):
        yield {
            'selected': self.value() == self.all_value,
            'query_string': cl.get_query_string({self.parameter_name: self.all_value}, []),
            'display': _('All'),
        }
        for lookup, title in self.lookup_choices:
            yield {
                'selected': self.value() == force_text(lookup) or (self.value() == None and force_text(self.default_value()) == force_text(lookup)),
                'query_string': cl.get_query_string({
                    self.parameter_name: lookup,
                }, []),
                'display': title,
            }

class StatusFilter(DefaultListFilter):
    title = _('Status ')
    parameter_name = 'status__exact'

    def lookups(self, request, model_admin):
        return ((0,'activate'), (1,'pending'), (2,'rejected'))

    def default_value(self):
        return 1

class MyModelAdmin(admin.ModelAdmin):
    list_filter = (StatusFilter,)

Here is my generic solution using redirect, it just checks if there are any GET parameters, if none exist then it redirects with the default get parameter. I also have a list_filter set so it picks that up and displays the default.

from django.shortcuts import redirect

class MyModelAdmin(admin.ModelAdmin):   

    ...

    list_filter = ('status', )

    def changelist_view(self, request, extra_context=None):
        referrer = request.META.get('HTTP_REFERER', '')
        get_param = "status__exact=5"
        if len(request.GET) == 0 and '?' not in referrer:
            return redirect("{url}?{get_parms}".format(url=request.path, get_parms=get_param))
        return super(MyModelAdmin,self).changelist_view(request, extra_context=extra_context)

The only caveat is when you do a direct get to the page with “?” present in the url, there is no HTTP_REFERER set so it will use the default parameter and redirect. This is fine for me, it works great when you click through the admin filter.

UPDATE:

In order to get around the caveat, I ended up writing a custom filter function which simplified the changelist_view functionality. Here is the filter:

class MyModelStatusFilter(admin.SimpleListFilter):
    title = _('Status')
    parameter_name = 'status'

    def lookups(self, request, model_admin):  # Available Values / Status Codes etc..
        return (
            (8, _('All')),
            (0, _('Incomplete')),
            (5, _('Pending')),
            (6, _('Selected')),
            (7, _('Accepted')),
        )

    def choices(self, cl):  # Overwrite this method to prevent the default "All"
        from django.utils.encoding import force_text
        for lookup, title in self.lookup_choices:
            yield {
                'selected': self.value() == force_text(lookup),
                'query_string': cl.get_query_string({
                    self.parameter_name: lookup,
                }, []),
                'display': title,
            }

    def queryset(self, request, queryset):  # Run the queryset based on your lookup values
        if self.value() is None:
            return queryset.filter(status=5)
        elif int(self.value()) == 0:
            return queryset.filter(status__lte=4)
        elif int(self.value()) == 8:
            return queryset.all()
        elif int(self.value()) >= 5:
            return queryset.filter(status=self.value())
        return queryset.filter(status=5)

And the changelist_view now only passes the default parameter if none are present. The idea was to get rid of the generics filters capability to view all by using no get parameters. To view all I assigned the status = 8 for that purpose.:

class MyModelAdmin(admin.ModelAdmin):   

    ...

    list_filter = ('status', )

    def changelist_view(self, request, extra_context=None):
        if len(request.GET) == 0:
            get_param = "status=5"
            return redirect("{url}?{get_parms}".format(url=request.path, get_parms=get_param))
        return super(MyModelAdmin, self).changelist_view(request, extra_context=extra_context)

def changelist_view( self, request, extra_context = None ):
    default_filter = False
    try:
        ref = request.META['HTTP_REFERER']
        pinfo = request.META['PATH_INFO']
        qstr = ref.split( pinfo )

        if len( qstr ) < 2:
            default_filter = True
    except:
        default_filter = True

    if default_filter:
        q = request.GET.copy()
        q['registered__exact'] = '1'
        request.GET = q
        request.META['QUERY_STRING'] = request.GET.urlencode()

    return super( InterestAdmin, self ).changelist_view( request, extra_context = extra_context )

You can simply usereturn queryset.filter() or if self.value() is None and Override method of SimpleListFilter

from django.utils.encoding import force_text

def choices(self, changelist):
    for lookup, title in self.lookup_choices:
        yield {
            'selected': force_text(self.value()) == force_text(lookup),
            'query_string': changelist.get_query_string(
                {self.parameter_name: lookup}, []
            ),
            'display': title,
        }

Note that if instead of pre-selecting a filter value you want to always pre-filter the data before showing it in the admin, you should override the ModelAdmin.queryset() method instead.

A slight improvement on Greg’s answer using DjangoChoices, Python >= 2.5 and of course Django >= 1.4.

from django.utils.translation import ugettext_lazy as _
from django.contrib.admin import SimpleListFilter

class OrderStatusFilter(SimpleListFilter):
    title = _('Status')

    parameter_name = 'status__exact'
    default_status = OrderStatuses.closed

    def lookups(self, request, model_admin):
        return (('all', _('All')),) + OrderStatuses.choices

    def choices(self, cl):
        for lookup, title in self.lookup_choices:
            yield {
                'selected': self.value() == lookup if self.value() else lookup == self.default_status,
                'query_string': cl.get_query_string({self.parameter_name: lookup}, []),
                'display': title,
            }

    def queryset(self, request, queryset):
        if self.value() in OrderStatuses.values:
            return queryset.filter(status=self.value())
        elif self.value() is None:
            return queryset.filter(status=self.default_status)


class Admin(admin.ModelAdmin):
    list_filter = [OrderStatusFilter] 

Thanks to Greg for the nice solution!

I know that is not the best solution, but i changed the index.html in the admin template, line 25 and 37 like this:

25: <th scope="row"><a href="{{ model.admin_url }}{% ifequal model.name "yourmodelname" %}?yourflag_flag__exact=1{% endifequal %}">{{ model.name }}</a></th>

37: <td><a href="{{ model.admin_url }}{% ifequal model.name "yourmodelname" %}?yourflag__exact=1{% endifequal %}" class="changelink">{% trans 'Change' %}</a></td>

I had to make a modification to get filtering to work correctly. The previous solution worked for me when the page loaded. If an ‘action’ was performed, the filter went back to ‘All’ and not my default. This solution loads the admin change page with the default filter, but also maintains filter changes or the current filter when other activity occurs on the page. I haven’t tested all cases, but in reality it may be limiting the setting of a default filter to occur only when the page loads.

def changelist_view(self, request, extra_context=None):
    default_filter = False

    try:
        ref = request.META['HTTP_REFERER']
        pinfo = request.META['PATH_INFO']
        qstr = ref.split(pinfo)
        querystr = request.META['QUERY_STRING']

        # Check the QUERY_STRING value, otherwise when
        # trying to filter the filter gets reset below
        if querystr is None:
            if len(qstr) < 2 or qstr[1] == '':
                default_filter = True
    except:
        default_filter = True

    if default_filter:
        q = request.GET.copy()
        q['registered__isnull'] = 'True'
        request.GET = q
        request.META['QUERY_STRING'] = request.GET.urlencode()

    return super(MyAdmin, self).changelist_view(request, extra_context=extra_context)

A bit off-topic but my search for a similar question led me here. I was looking to have a default query by a date (ie if no input is provided, show only objects with timestamp of ‘Today’), which complicates the question a bit. Here is what I came up with:

from django.contrib.admin.options import IncorrectLookupParameters
from django.core.exceptions import ValidationError

class TodayDefaultDateFieldListFilter(admin.DateFieldListFilter):
    """ If no date is query params are provided, query for Today """

    def queryset(self, request, queryset):
        try:
            if not self.used_parameters:
                now = datetime.datetime.now().replace(hour=0, minute=0, second=0, microsecond=0)
                self.used_parameters = {
                    ('%s__lt' % self.field_path): str(now + datetime.timedelta(days=1)),
                    ('%s__gte' % self.field_path): str(now),
                }
                # Insure that the dropdown reflects 'Today'
                self.date_params = self.used_parameters
            return queryset.filter(**self.used_parameters)
        except ValidationError, e:
            raise IncorrectLookupParameters(e)

class ImagesAdmin(admin.ModelAdmin):
    list_filter = (
        ('timestamp', TodayDefaultDateFieldListFilter),
    )

This is a simple override of the default DateFieldListFilter. By setting self.date_params, it insures that the filter dropdown will update to whatever option matches the self.used_parameters. For this reason, you must insure that the self.used_parameters are exactly what would be used by one of those dropdown selections (ie, find out what the date_params would be when using the ‘Today’ or ‘Last 7 Days’ and construct the self.used_parameters to match those).

This was built to work with Django 1.4.10

This may be an old thread, but thought I would add my solution as I couldn’t find better answers on google searches.

Do what (not sure if its Deminic Rodger, or ha22109) answered in the ModelAdmin for changelist_view

class MyModelAdmin(admin.ModelAdmin):   
    list_filter = (CustomFilter,)

    def changelist_view(self, request, extra_context=None):

        if not request.GET.has_key('decommissioned__exact'):

            q = request.GET.copy()
            q['decommissioned__exact'] = 'N'
            request.GET = q
            request.META['QUERY_STRING'] = request.GET.urlencode()
        return super(MyModelAdmin,self).changelist_view(request, extra_context=extra_context)

Then we need to create a custom SimpleListFilter

class CustomFilter(admin.SimpleListFilter):
    title = 'Decommissioned'
    parameter_name = 'decommissioned'  # i chose to change it

def lookups(self, request, model_admin):
    return (
        ('All', 'all'),
        ('1', 'Decommissioned'),
        ('0', 'Active (or whatever)'),
    )

# had to override so that we could remove the default 'All' option
# that won't work with our default filter in the ModelAdmin class
def choices(self, cl):
    yield {
        'selected': self.value() is None,
        'query_string': cl.get_query_string({}, [self.parameter_name]),
        # 'display': _('All'),
    }
    for lookup, title in self.lookup_choices:
        yield {
            'selected': self.value() == lookup,
            'query_string': cl.get_query_string({
                self.parameter_name: lookup,
            }, []),
            'display': title,
        }

def queryset(self, request, queryset):
    if self.value() == '1':
        return queryset.filter(decommissioned=1)
    elif self.value() == '0':
        return queryset.filter(decommissioned=0)
    return queryset

Here’s the Cleanest version I was able to generate of a filter with a redefined ‘All’ and a Default value that is selected.

If shows me by default the Trips currently happening.

class HappeningTripFilter(admin.SimpleListFilter):
    """
    Filter the Trips Happening in the Past, Future or now.
    """
    default_value = 'now'
    title = 'Happening'
    parameter_name = 'happening'

    def lookups(self, request, model_admin):
        """
        List the Choices available for this filter.
        """
        return (
            ('all', 'All'),
            ('future', 'Not yet started'),
            ('now', 'Happening now'),
            ('past', 'Already finished'),
        )

    def choices(self, changelist):
        """
        Overwrite this method to prevent the default "All".
        """
        value = self.value() or self.default_value
        for lookup, title in self.lookup_choices:
            yield {
                'selected': value == force_text(lookup),
                'query_string': changelist.get_query_string({
                    self.parameter_name: lookup,
                }, []),
                'display': title,
            }

    def queryset(self, request, queryset):
        """
        Returns the Queryset depending on the Choice.
        """
        value = self.value() or self.default_value
        now = timezone.now()
        if value == 'future':
            return queryset.filter(start_date_time__gt=now)
        if value == 'now':
            return queryset.filter(start_date_time__lte=now, end_date_time__gte=now)
        if value == 'past':
            return queryset.filter(end_date_time__lt=now)
        return queryset.all()

Created a reusable Filter sub-class, inspired by some of the answers here (mostly Greg’s).

Advantages:

Reusable – Pluggable in any standard ModelAdmin classes

Extendable – Easy to add additional/custom logic for QuerySet filtering

Easy to use – In its most basic form, only one custom attribute and one custom method need to be implemented (apart from those required for SimpleListFilter subclassing)

Intuitive admin – The “All” filter link is working as expected; as are all the others

No redirects – No need to inspect GET request payload, agnostic of HTTP_REFERER (or any other request related stuff, in its basic form)

No (changelist) view manipulation – And no template manipulations (god forbid)

Code:

(most of the imports are just for type hints and exceptions)

from typing import List, Tuple, Any

from django.contrib.admin.filters import SimpleListFilter
from django.contrib.admin.options import IncorrectLookupParameters
from django.contrib.admin.views.main import ChangeList
from django.db.models.query import QuerySet
from django.utils.encoding import force_str
from django.utils.translation import gettext_lazy as _
from django.core.exceptions import ValidationError


class PreFilteredListFilter(SimpleListFilter):

    # Either set this or override .get_default_value()
    default_value = None

    no_filter_value = 'all'
    no_filter_name = _("All")

    # Human-readable title which will be displayed in the
    # right admin sidebar just above the filter options.
    title = None

    # Parameter for the filter that will be used in the URL query.
    parameter_name = None

    def get_default_value(self):
        if self.default_value is not None:
            return self.default_value
        raise NotImplementedError(
            'Either the .default_value attribute needs to be set or '
            'the .get_default_value() method must be overridden to '
            'return a URL query argument for parameter_name.'
        )

    def get_lookups(self) -> List[Tuple[Any, str]]:
        """
        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.
        """
        raise NotImplementedError(
            'The .get_lookups() method must be overridden to '
            'return a list of tuples (value, verbose value).'
        )

    # Overriding parent class:
    def lookups(self, request, model_admin) -> List[Tuple[Any, str]]:
        return [(self.no_filter_value, self.no_filter_name)] + self.get_lookups()

    # Overriding parent class:
    def queryset(self, request, queryset: QuerySet) -> QuerySet:
        """
        Returns the filtered queryset based on the value
        provided in the query string and retrievable via
        `self.value()`.
        """
        if self.value() is None:
            return self.get_default_queryset(queryset)
        if self.value() == self.no_filter_value:
            return queryset.all()
        return self.get_filtered_queryset(queryset)

    def get_default_queryset(self, queryset: QuerySet) -> QuerySet:
        return queryset.filter(**{self.parameter_name: self.get_default_value()})

    def get_filtered_queryset(self, queryset: QuerySet) -> QuerySet:
        try:
            return queryset.filter(**self.used_parameters)
        except (ValueError, ValidationError) as e:
            # Fields may raise a ValueError or ValidationError when converting
            # the parameters to the correct type.
            raise IncorrectLookupParameters(e)

    # Overriding parent class:
    def choices(self, changelist: ChangeList):
        """
        Overridden to prevent the default "All".
        """
        value = self.value() or force_str(self.get_default_value())
        for lookup, title in self.lookup_choices:
            yield {
                'selected': value == force_str(lookup),
                'query_string': changelist.get_query_string({self.parameter_name: lookup}),
                'display': title,
            }

Full usage example:

from django.contrib import admin
from .models import SomeModelWithStatus


class StatusFilter(PreFilteredListFilter):
    default_value = SomeModelWithStatus.Status.FOO
    title = _('Status')
    parameter_name = 'status'

    def get_lookups(self):
        return SomeModelWithStatus.Status.choices


@admin.register(SomeModelWithStatus)
class SomeModelAdmin(admin.ModelAdmin):
    list_filter = (StatusFilter, )

Hope this helps somebody; feedback always appreciated.

How can I create more than one ModelAdmin for the same model, each customised differently and linked to different URLs?

Let’s say I have a Django model called Posts. By default, the admin view of this model will list all Post objects.

I know I can customise the list of objects displayed on the page in various ways by setting variables like list_display or overriding the queryset method in my ModelAdmin like so:

class MyPostAdmin(admin.ModelAdmin):
    list_display = ('title', 'pub_date')

    def queryset(self, request):
        request_user = request.user
        return Post.objects.filter(author=request_user)

admin.site.register(MyPostAdmin, Post)

By default, this would be accessible at the URL /admin/myapp/post. However I would like to have multiple views/ModelAdmins of the same model. e.g /admin/myapp/post would list all post objects, and /admin/myapp/myposts would list all posts belonging to the user, and /admin/myapp/draftpost might list all posts that have not yet been published. (these are just examples, my actual use-case is more complex)

You cannot register more than one ModelAdmin for the same model (this results in an AlreadyRegistered exception). Ideally I’d like to achieve this without putting everything into a single ModelAdmin class and writing my own ‘urls’ function to return a different queryset depending on the URL.

I’ve had a look at the Django source and I see functions like ModelAdmin.changelist_view that could be somehow included in my urls.py, but I’m not sure exactly how that would work.

Update: I’ve found one way of doing what I want (see below), but I’d still like to hear other ways of doing this.

I’ve found one way to achieve what I want, by using proxy models to get around the fact that each model may be registered only once.

class PostAdmin(admin.ModelAdmin):
    list_display = ('title', 'pubdate','user')

class MyPost(Post):
    class Meta:
        proxy = True

class MyPostAdmin(PostAdmin):
    def get_queryset(self, request):
        return self.model.objects.filter(user = request.user)


admin.site.register(Post, PostAdmin)
admin.site.register(MyPost, MyPostAdmin)

Then the default PostAdmin would be accessible at /admin/myapp/post and the list of posts owned by the user would be at /admin/myapp/myposts.

After looking at http://code.djangoproject.com/wiki/DynamicModels, I’ve come up with the following function utility function to do the same thing:

def create_modeladmin(modeladmin, model, name = None):
    class  Meta:
        proxy = True
        app_label = model._meta.app_label

    attrs = {'__module__': '', 'Meta': Meta}

    newmodel = type(name, (model,), attrs)

    admin.site.register(newmodel, modeladmin)
    return modeladmin

This can be used as follows:

class MyPostAdmin(PostAdmin):
    def get_queryset(self, request):
        return self.model.objects.filter(user = request.user)

create_modeladmin(MyPostAdmin, name='my-posts', model=Post)

Paul Stone answer is absolutely great! Just to add, for Django 1.4.5 I needed to inherit my custom class from admin.ModelAdmin

class MyPostAdmin(admin.ModelAdmin):
    def queryset(self, request):
        return self.model.objects.filter(id=1)

I have a django site with lots of models and forms. I have many custom forms and formsets and inlineformsets and custom validation and custom querysets. Hence the add model action depends on forms that need other things, and the ‘add model’ in the django admin throughs a 500 from a custom queryset.

Is there anyway to disable the ‘Add $MODEL’ functionality for a certain models?

I want /admin/appname/modelname/add/ to give a 404 (or suitable ‘go away’ error message), I don’t want the ‘Add $MODELNAME’ button to be on /admin/appname/modelname view.

Django admin provides a way to disable admin actions (http://docs.djangoproject.com/en/dev/ref/contrib/admin/actions/#disabling-actions) however the only action for this model is ‘delete_selected’. i.e. the admin actions only act on existing models. Is there some django-esque way to do this?

It is easy, just overload has_add_permission method in your Admin class like so:

class MyAdmin(admin.ModelAdmin):
     def has_add_permission(self, request, obj=None):
        return False

By default syncdb creates 3 security permissions for each model:

1. Create (aka add)
2. Change
3. Delete

If your logged in as Admin, you get EVERYTHING no matter what.

But if you create a new user group called “General Access” (for example) then you can assign ONLY the CHANGE and DELETE permissions for all of your models.

Then any logged in user that is a member of that group will not have “Create” permission, nothing related to it will show on the screen.

I think this will help you.. below code must be in admin.py file

@admin.register(Author)
class AuthorAdmin(admin.ModelAdmin):
    list_display = ('name', )
    list_filter = ('name', )
    search_fields = ('name', )
    list_per_page = 20

    # This will help you to disbale add functionality
    def has_add_permission(self, request):
        return False

    # This will help you to disable delete functionaliyt
    def has_delete_permission(self, request, obj=None):
        return False

In additon to the above as posted by

    # This will help you to disable change functionality
    def has_change_permission(self, request, obj=None):
        return False

Just copy code from another answer

# In admin
# make the related field can't be added
    def get_form(self, request, obj=None, **kwargs):
        form = super().get_form(request, obj, **kwargs)
        form.base_fields['service'].widget.can_add_related = False
        return form

In my case I use inline

# In inline formset e.g. admin.TabularInline
# disable all
    def get_formset(self, request, obj=None, **kwargs):
        formset = super().get_formset(request, obj, **kwargs)
        service = formset.form.base_fields['service']
        service.widget.can_add_related = service.widget.can_change_related = service.widget.can_delete_related = False
        return formset

in service = formset.form.base_fields['service'] base_fields is the fields defined in model

if defined in the form use:

product = formset.form.declared_fields['product']

see also

This is a too much delayed answer; Just posting this as if anyone is finding the same solution.

In admin.py file you can do the following:

class MyModelForm(forms.ModelForm):

class Meta:
    model = MyModel
    fields = '__all__'


class MyModelAdmin(admin.ModelAdmin):
    form = QuestionTrackAdminForm
    list_display = ['title', 'weight']
    readonly_fields = ['title', 'weight']

admin.site.register(MyModel, MyModelAdmin)

Here, “readonly_fields” does the magic. Thanks.

How does one change the ‘Django administration’ text in the django admin header?

It doesn’t seem to be covered in the “Customizing the admin” documentation.

Update: If you are using Django 1.7+, see the answer below.

Original answer from 2011: You need to create your own admin base_site.html template to do this. The easiest way is to create the file:

/<projectdir>/templates/admin/base_site.html

This should be a copy of the original base_site.html, except putting in your custom title:

{% block branding %}
<h1 id="site-name">{% trans 'my cool admin console' %}</h1>
{% endblock %}

For this to work, you need to have the correct settings for your project, namely in settings.py:

• Make sure /projectdir/templates/ is added into TEMPLATE_DIRS.
• Make sure django.template.loaders.filesystem.Loader is added into TEMPLATE_LOADERS.

See docs for more information on settings.py.

As of Django 1.7 you don’t need to override templates. You can now implement site_header, site_title, and index_title attributes on a custom AdminSite in order to easily change the admin site’s page title and header text. Create an AdminSite subclass and hook your instance into your URLconf:

admin.py:

from django.contrib.admin import AdminSite
from django.utils.translation import ugettext_lazy

class MyAdminSite(AdminSite):
    # Text to put at the end of each page's <title>.
    site_title = ugettext_lazy('My site admin')

    # Text to put in each page's <h1> (and above login form).
    site_header = ugettext_lazy('My administration')

    # Text to put at the top of the admin index page.
    index_title = ugettext_lazy('Site administration')

admin_site = MyAdminSite()

urls.py:

from django.conf.urls import patterns, include
from myproject.admin import admin_site

urlpatterns = patterns('',
    (r'^myadmin/', include(admin_site.urls)),
)

Update: As pointed out by oxfn you can simply set the site_header in your urls.py or admin.py directly without subclassing AdminSite:

admin.site.site_header = 'My administration'

There is an easy way to set admin site header – assign it to current admin instance in urls.py like this

admin.site.site_header = 'My admin'

Or one can implement some header-building magic in separate method

admin.site.site_header = get_admin_header()

Thus, in simple cases there’s no need to subclass AdminSite

In urls.py you can override the 3 most important variables:

from django.contrib import admin

admin.site.site_header = 'My project'                    # default: "Django Administration"
admin.site.index_title = 'Features area'                 # default: "Site administration"
admin.site.site_title = 'HTML title from adminsitration' # default: "Django site admin"

Reference: Django documentation on these attributes.

A simple complete solution in Django 1.8.3 based on answers in this question.

In settings.py add:

ADMIN_SITE_HEADER = "My shiny new administration"

In urls.py add:

from django.conf import settings
admin.site.site_header = settings.ADMIN_SITE_HEADER

The easiest way of doing it make sure you have

from django.contrib import admin

and then just add these at bottom of url.py of you main application

admin.site.site_title = "Your App Title"
admin.site.site_header = "Your App Admin" 

For Django 2.1.1 add following lines to urls.py

from django.contrib import admin

# Admin Site Config
admin.sites.AdminSite.site_header = 'My site admin header'
admin.sites.AdminSite.site_title = 'My site admin title'
admin.sites.AdminSite.index_title = 'My site admin index'

As you can see in the templates, the text is delivered via the localization framework (note the use of the trans template tag). You can make changes to the translation files to override the text without making your own copy of the templates.

1. mkdir locale

2. ./manage.py makemessages

3. Edit locale/en/LC_MESSAGES/django.po, adding these lines:

   msgid "Django site admin"
   msgstr "MySite site admin"
   
   msgid "Django administration"
   msgstr "MySite administration"
   

4. ./manage.py compilemessages

See https://docs.djangoproject.com/en/1.3/topics/i18n/localization/#message-files

admin.py:

from django.contrib.admin import AdminSite

AdminSite.site_title = ugettext_lazy('My Admin')

AdminSite.site_header = ugettext_lazy('My Administration')

AdminSite.index_title = ugettext_lazy('DATA BASE ADMINISTRATION')

First of all, you should add templates/admin/base_site.html to your project. This file can safely be overwritten since it’s a file that the Django devs have intended for the exact purpose of customizing your admin site a bit. Here’s an example of what to put in the file:

{% extends "admin/base.html" %}
{% load i18n %}

{% block title %}{{ title }} | {% trans 'Some Organisation' %}{% endblock %}

{% block branding %}
<style type="text/css">
  #header
  {
    /* your style here */
  }
</style>
<h1 id="site-name">{% trans 'Organisation Website' %}</h1>
{% endblock %}

{% block nav-global %}{% endblock %}

This is common practice. But I noticed after this that I was still left with an annoying “Site Administration” on the main admin index page. And this string was not inside any of the templates, but rather set inside the admin view. Luckily it’s quite easy to change. Assuming your language is set to English, run the following commands from your project directory:

$ mkdir locale
$ ./manage.py makemessages -l en

Now open up the file locale/en/LC_MESSAGES/django.po and add two lines after the header information (the last two lines of this example)

"Project-Id-Version: PACKAGE VERSION\n"
"Report-Msgid-Bugs-To: \n"
"POT-Creation-Date: 2010-04-03 03:25+0200\n"
"PO-Revision-Date: YEAR-MO-DA HO:MI+ZONE\n"
"Last-Translator: FULL NAME <EMAIL@ADDRESS>\n"
"Language-Team: LANGUAGE <LL@li.org>\n"
"MIME-Version: 1.0\n"
"Content-Type: text/plain; charset=UTF-8\n"
"Content-Transfer-Encoding: 8bit\n"

msgid "Site administration"
msgstr "Main administration index"

After this, remember to run the following command and reload your project’s server:

$ ./manage.py compilemessages

source: http://overtag.dk/wordpress/2010/04/changing-the-django-admin-site-title/