Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

How to override and extend basic Django admin templates?

People also ask

How do I override a Django template?

Extending an overridden template With your template loaders configured, you can extend a template using the {% extends %} template tag whilst at the same time overriding it. This can allow you to make small customizations without needing to reimplement the entire template.

Can we change Django admin template?

To do so, you will have to change the project's settings.py . Find the TEMPLATES section and modify accordingly. To override the default template you first need to access the template you want to modify from the django/contrib/admin/templates/admin directory.

How customize Django admin UI?

1. Change model name: If you want to change name of model which is States here so open model.py file and add verbose_name attribute in meta section.

How do I change the admin text in Django?

To change the admin site header text, login page, and the HTML title tag of our bookstore's instead, add the following code in urls.py . The site_header changes the Django administration text which appears on the login page and the admin site. The site_title changes the text added to the <title> of every admin page.


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.

  1. 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',
                ],
            },
        },
    ]
    
  2. 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.

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

    {% extends "admin/change_form.html" %}
    {% block field_sets %}
         {# your modification here #}
    {% endblock %}
    
  2. 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.


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.


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.