Below is my model
class Movie(models.Model):
name = models.CharField(max_length=55)
artists = models.ManyToManyField(Artist)
def __str__(self):
return self.name
Below is the View:
def moviesView(request):
movies = Movie.objects.all()
context = {
"movies": movies
}
return render(request, 'movies/movies.html', context=context)
def movieView(request, name):
print(name)
movie = Movie.object.get(name=name)
context = {
"movie": movie
}
return render(request, 'movies/movie.html', context=context)
Below are the URLS:
urlpatterns = [
path('movies', moviesView, name='movies'),
re_path('movies/(\d+)', movieView, name='movie'),
path('artists', artistView, name='artists')
]
Below is the template:
<h1>Movies</h1>
{% for movie in movies %}
{{ movie.name }}
<h3>{{ movie.name }}</h3>
{% for artist in movie.artists.all %}
<ul>
<li>{{ artist.name }}</li>
</ul>
{% endfor %}
{% endfor %}
If I click a movie avengers, it should carry to another page with movie details of avengers as mentioned in the model:
I need to frame the url as: http://127.0.0.1:8000/movies/avengers
You did not specify a parameter. But even if you did, it would not accept it, since \d+ only accepts a sequence of digits, not strings.
You can work with:
urlpatterns = [
path('movies/', moviesView, name='movies'),
path('movies/(?P<name>.*)/', movieView, name='movie'),
path('artists/', artistView, name='artists')
]
but it is likely easier to work with path instead:
urlpatterns = [
path('movies/', moviesView, name='movies'),
re_path('movies/<str:name>/', movieView, name='movie'),
path('artists/', artistView, name='artists')
]
It is also not a good idea to work with a name. A lot of characters will be percent encoded [wiki] resulting in ugly URLs. Django uses slugs to make visually pleasant URLs.
Update :: Problem solved.just follow the guy below.
in my urls.py
path('', store_view, name='store'),
path('category/<str:category_name>/', category_view, name='category'),
in views.py
def store_view(request):
categories = list(Category.objects.all())
context = {
'categories': categories,
}
return render(request, 'store/store.html', context)
def category_view(request, category_name):
category = Category.objects.get(name=category_name)
context = {
'category': category,
}
return render(request, 'store/single-category-view.html', context)
in my template : store.html , that is rendered by store_view >>
{% for item in categories %}
<a href="{% url 'category' item.name %}">
{{item.name}}
</a>
{% endfor %}
Now,the problem is, in the category column in my DB, i have got one category called 'Laptop/MacBook'.when ever this name is passed to the url, it says >>
"Reverse for 'category' with arguments '('Laptop/MacBook',)' not
found. 1 pattern(s) tried: ['category/(?P<category_name>[^/]+)/$']
But when i changed the category name from Laptop/MacBook to Laptop and MacBook , it worked fine and showed no error.
But i want to keep it as it was,'Laptop/MacBook'.How can i do that??and how do you guys deal with that?
Try encoding and decoding your DB values. Assuming its Python 3:
from urllib.parse import quote, unquote
encoded = quote("Laptop/Macbook", safe="")
decoded = unquote(encoded)
print(encoded, decoded)
Output:
Laptop%2FMacbook Laptop/Macbook
With this your route should take in the right param.
from django.http import HttpResponse, request
from django.shortcuts import render
def store_view(request):
name = "Laptop/Macbook"
return render(request, './store.html', context={"name": name})
def category_view(request, category_name):
print(category_name)
return HttpResponse(b"Here we go!")
templatetags/tags.py
from urllib.parse import quote, unquote
from django import template
register = template.Library()
#register.filter(name='encode')
def encode(name):
return quote(name, safe="")
#register.filter(name='decode')
def decode(name):
return unquote(name)
Template:
{% load tags %}
<a href="{% url 'category' name|encode %}">
{{name}}
</a>
Don't forget to add in settings:
'OPTIONS': {
'libraries':{
'tags': 'templatetags.tags',
}
},
When using a "/", django thinks that you are passing more than one parameter. To fix this, replace str by path in your urls like so:
path('', store_view, name='store'),
path('category/<path:category_name>/', category_view, name='category'),
This will make django understand that the / does not mean there are two separate parameters in your url.
I'm new to Django and slowly learning how it works. I just upgraded to 1.10 and part of my app stopped working. I know it is related to the changes made into Reverse. I have been reading and I cannot find exactly what I'm doing wrong. Almost everything works as it should with a couple of exceptions. The behavior is as follows:
1) On my app I load reservations/create, it works perfectly I can create my reservation
2) When I click create, the reservation is actually created and saved into the database, but the browser is sent to the wrong address. It gets sent to reservations/create instead of reservations/reservation number (for example reservations/2 where it shows its details) and shows a Reverse error (included in this post)
3) If I test reservations/2 for example, it shows that it was actually created.
4) Also if a go straight to reservations/ it should show a list of all the ones already create, but instead shows a Reverse error too.
I would really appreciate any help in understanding what I'm doing wrong.
Models.py
class Reservation(models.Model):
res_number = models.AutoField(primary_key=True)
date = models.DateField(default=datetime.date.today())
status = models.CharField(max_length=10,default="Created")
reservation_type = models.CharField(max_length=11,choices=shced_type_choices, default="rental")
aircraft = models.ForeignKey('aircraft.Aircraft')
renter = models.CharField(max_length=30,blank=False,null=False)
instructor = models.CharField(max_length=30,blank=True,null=False)
def get_absolute_url(self):
return reverse("reservations:detail", kwargs={"res_number": self.res_number})
Main urls.py
url(r'^reservations/', include('dispatch.urls', namespace='reservations')),
Dispatch.urls
from django.conf.urls import include, url
from django.contrib import admin
from .views import (
reservations_list,
reservations_detail,
reservations_edit,
reservations_dispatch,
reservations_close,
reservations_cancel,
reservations_create,
reservations_close,
)
urlpatterns = [
url(r'^$', reservations_list),
url(r'^(?P<res_number>\d+)/$', reservations_detail),
url(r'^(?P<res_number>\d+)/edit/$', reservations_edit),
url(r'^(?P<res_number>\d+)/dispatch/$', reservations_dispatch),
url(r'^(?P<res_number>\d+)/close/$', reservations_close),
url(r'^(?P<res_number>\d+)/cancel/$', reservations_cancel),
url(r'^create/$', reservations_create),
url(r'^close/$', reservations_close),
]
Views.py
from django.contrib import messages
from django import forms
from django.http import HttpResponse, HttpResponseRedirect
from django.shortcuts import render, get_object_or_404
from .forms import ReservationForm, CloseReservationForm
from .models import Reservation
def reservations_list(request):
queryset = Reservation.objects.all()
context = {
"object_list": queryset,
"title": "List of Reservations:"
}
return render(request, "dispatch/list.html", context)
def reservations_detail(request, res_number=None):
instance = get_object_or_404(Reservation, res_number=res_number)
context = {
"title": instance.renter,
"instance": instance,
}
return render(request, "dispatch/details.html", context)
def reservations_create(request):
form = ReservationForm(request.POST or None)
if form.is_valid():
instance = form.save(commit=False)
print(instance.aircraft.hobbs)
instance.save()
messages.success(request, "Reservation Created")
return HttpResponseRedirect(instance.get_absolute_url())
context = {
"form": form,
}
return render(request, "dispatch/create.html", context)
Details.html
{% extends "dispatch/base.html" %}
{% block head_title %}{{ block.super }} | {{instance.res_number}}{% endblock head_title %}
{% block content %}
<h1>Reservation for {{title}} on {{instance.date}}</h1>
Reservation Number: {{instance.res_number}}</br>
Date: {{instance.date}}</br>
Status: {{instance.status}}</br>
Reservation Type: {{instance.reservation_type}}</br>
Aircraft: {{instance.aircraft}}</br>
Renter's Name: {{instance.renter}}</br>
Instructor's Name: {{instance.instructor}}</br>
Expected Flight Hours: {{instance.expected_hours}} Hrs</br>
Actual Flown Hours: {{instance.flown_hours}} Hrs</br>
Reservation Created on: {{instance.created}}</br>
Last Updated on: {{instance.updated}}</br>
{% endblock content %}
Create.html
{% extends "dispatch/base.html" %}
{% block head_title %}{{ block.super }} | Create{% endblock head_title %}
{% block content %}
<h1>Create Reservation</h1>
<form method='POST' action=''>{% csrf_token %}
{{form.as_p}}
<input type="submit" name="Create Reservation">
</form>
{% endblock content %}
Reverse error screenshot
Your problem is your routes don't have names. So when you are using reverse('some_name'), you have to have such name defined. The name is detail in your case, so you want to do something like this (see the parameter name)
urlpatterns = [
url(r'^(?P<res_number>\d+)/$', reservations_detail, name='detail'),
]
Also please don't insert traceback as a screenshot. You see the link 'switch to copy-and-paste view'? Yeah, use that the next time.
Let's say I have my pizza application with Topping and Pizza classes and they show in Django Admin like this:
PizzaApp
-
Toppings >>>>>>>>>> Add / Change
Pizzas >>>>>>>>>> Add / Change
But I want them like this:
PizzaApp
-
Pizzas >>>>>>>>>> Add / Change
Toppings >>>>>>>>>> Add / Change
How do I configure that in my admin.py?
A workaround that you can try is tweaking your models.py as follows:
class Topping(models.Model):
.
.
.
class Meta:
verbose_name_plural = "2. Toppings"
class Pizza(models.Model):
.
.
.
class Meta:
verbose_name_plural = "1. Pizzas"
Not sure if it is against the django's best practices but it works (tested with django trunk).
Good luck!
PS: sorry if this answer was posted too late but it can help others in future similar situations.
If you want to solve this in 10 seconds just use spaces in verbose_name_plural, for example:
class Topping(models.Model):
class Meta:
verbose_name_plural = " Toppings" # 2 spaces
class Pizza(models.Model):
class Meta:
verbose_name_plural = " Pizzas" # 1 space
Of course it isn't ellegant but may work for a while before we get a better solution.
I eventually managed to do it thanks to this Django snippet, you just need to be aware of the ADMIN_REORDER setting:
ADMIN_REORDER = (
('app1', ('App1Model1', 'App1Model2', 'App1Model3')),
('app2', ('App2Model1', 'App2Model2')),
)
app1 must not be prefixed with the project name, i.e. use app1 instead of mysite.app1.
There's now a nice Django package for that:
https://pypi.python.org/pypi/django-modeladmin-reorder
Answer in June 2018
This answer is similar to Vasil's idea
I tried to solve similar problems, and then I saw such the fragment.
I made some modifications based on this clip. The code is as follows.
# myproject/setting.py
...
# set my ordering list
ADMIN_ORDERING = [
('pizza_app', [
'Pizzas',
'Toppings'
]),
]
# Creating a sort function
def get_app_list(self, request):
app_dict = self._build_app_dict(request)
for app_name, object_list in ADMIN_ORDERING:
app = app_dict[app_name]
app['models'].sort(key=lambda x: object_list.index(x['object_name']))
yield app
# Covering django.contrib.admin.AdminSite.get_app_list
from django.contrib import admin
admin.AdminSite.get_app_list = get_app_list
...
Note that the sorting list used in this sorting function contains the sorting of all app and its modules in the system. If you don't need it, please design the sorting function according to your own needs.
It works great on Django 2.0
This is actually covered at the very bottom of Writing your first Django app, part 7.
Here's the relevant section:
Customize the admin index page
On a similar note, you might want to
customize the look and feel of the
Django admin index page.
By default, it displays all the apps
in INSTALLED_APPS that have been
registered with the admin application,
in alphabetical order. You may want to
make significant changes to the
layout. After all, the index is
probably the most important page of
the admin, and it should be easy to
use.
The template to customize is
admin/index.html. (Do the same as with
admin/base_site.html in the previous
section -- copy it from the default
directory to your custom template
directory.) Edit the file, and you'll
see it uses a template variable called
app_list. That variable contains every
installed Django app. Instead of using
that, you can hard-code links to
object-specific admin pages in
whatever way you think is best.
Here's the snippet Emmanuel used, updated for Django 1.8:
In templatetags/admin_reorder.py:
from django import template
from django.conf import settings
from collections import OrderedDict
register = template.Library()
# from http://www.djangosnippets.org/snippets/1937/
def register_render_tag(renderer):
"""
Decorator that creates a template tag using the given renderer as the
render function for the template tag node - the render function takes two
arguments - the template context and the tag token
"""
def tag(parser, token):
class TagNode(template.Node):
def render(self, context):
return renderer(context, token)
return TagNode()
for copy_attr in ("__dict__", "__doc__", "__name__"):
setattr(tag, copy_attr, getattr(renderer, copy_attr))
return register.tag(tag)
#register_render_tag
def admin_reorder(context, token):
"""
Called in admin/base_site.html template override and applies custom ordering
of apps/models defined by settings.ADMIN_REORDER
"""
# sort key function - use index of item in order if exists, otherwise item
sort = lambda order, item: (order.index(item), "") if item in order else (
len(order), item)
if "app_list" in context:
# sort the app list
order = OrderedDict(settings.ADMIN_REORDER)
context["app_list"].sort(key=lambda app: sort(order.keys(),
app["app_url"].strip("/").split("/")[-1]))
for i, app in enumerate(context["app_list"]):
# sort the model list for each app
app_name = app["app_url"].strip("/").split("/")[-1]
if not app_name:
app_name = app["name"].lower()
model_order = [m.lower() for m in order.get(app_name, [])]
context["app_list"][i]["models"].sort(key=lambda model:
sort(model_order, model["admin_url"].strip("/").split("/")[-1]))
return ""
In settings.py:
ADMIN_REORDER = (
('app1', ('App1Model1', 'App1Model2', 'App1Model3')),
('app2', ('App2Model1', 'App2Model2')),
)
(insert your own app names in here. Admin will place missing apps or models at the end of the list, so long as you list at least two models in each app.)
In your copy of base_site.html:
{% extends "admin/base.html" %}
{% load i18n admin_reorder %}
{% block title %}{{ title }} | {% trans 'Django site admin' %}{% endblock %}
{% block branding %}
{% admin_reorder %}
<h1 id="site-name">{% trans 'Django administration' %}</h1>
{% endblock %}
{% block nav-global %}{% endblock %}
ADMIN_ORDERING = {
"PizzaApp": [
"Pizzas",
"Toppings",
],
}
def get_app_list(self, request):
app_dict = self._build_app_dict(request)
for app_name, object_list in app_dict.items():
if app_name in ADMIN_ORDERING:
app = app_dict[app_name]
app["models"].sort(
key=lambda x: ADMIN_ORDERING[app_name].index(x["object_name"])
)
app_dict[app_name]
yield app
else:
yield app_dict[app_name]
admin.AdminSite.get_app_list = get_app_list
This solution works for me, modified the one from ζδΌι.
You get to keep the default auth ordering AND specify your own.
If you're using Suit for the AdminSite you can do menu customization using the menu tag.
I was looking for a simple solution where I could order the apps by their name in the admin panel. I came up with the following template tag:
from django import template
from django.conf import settings
register = template.Library()
#register.filter
def sort_apps(apps):
apps.sort(
key = lambda x:
settings.APP_ORDER.index(x['app_label'])
if x['app_label'] in settings.APP_ORDER
else len(apps)
)
print [x['app_label'] for x in apps]
return apps
Then, just override templates/admin/index.html and add that template tag:
{% extends "admin/index.html" %}
{% block content %}
{% load i18n static sort_apps %}
<div id="content-main">
{% if app_list %}
{% for app in app_list|sort_apps %}
<div class="app-{{ app.app_label }} module">
<table>
<caption>
{{ app.name }}
</caption>
{% for model in app.models %}
<tr class="model-{{ model.object_name|lower }}">
{% if model.admin_url %}
<th scope="row">{{ model.name }}</th>
{% else %}
<th scope="row">{{ model.name }}</th>
{% endif %}
{% if model.add_url %}
<td>{% trans 'Add' %}</td>
{% else %}
<td> </td>
{% endif %}
{% if model.admin_url %}
<td>{% trans 'Change' %}</td>
{% else %}
<td> </td>
{% endif %}
</tr>
{% endfor %}
</table>
</div>
{% endfor %}
{% else %}
<p>{% trans "You don't have permission to edit anything." %}</p>
{% endif %}
</div>
{% endblock %}
Then customized the APP_ORDER in settings.py:
APP_ORDER = [
'app1',
'app2',
# and so on...
]
It works great on Django 1.10
Here's a version that gives you a bit more flexibility, namely:
You can partially define apps ordering, leaving the rest for Django to add to the list
You can specify order on modules, or avoid defining it, by using '*' instead
Your defined apps ordering will appear first, then all the rest of apps appended after it
To check the name of your app, either look at the file apps.py inside the app's directory and check for name property of the class Config(AppConfi): or in case that is not present, use the name of the directory for the app in the project.
Add this code somewhere in your settings.py file:
# ======[Setting the order in which the apps/modules show up listed on Admin]========
# set my ordering list
ADMIN_ORDERING = [
('crm', '*'),
('property', '*'),
]
# Creating a sort function
def get_app_list(self, request):
"""
Returns a sorted list of all the installed apps that have been
registered in this site.
Allows for:
ADMIN_ORDERING = [
('app_1', [
'module_1',
'module_2'
]),
('app_2', '*'),
]
"""
app_dict = self._build_app_dict(request)
# Let's start by sorting the apps alphabetically on a list:
app_list = sorted(app_dict.values(), key=lambda x: x['name'].lower())
# Sorting the models alphabetically within each app.
for app in app_list:
if app['app_label'] in [el[0] for el in ADMIN_ORDERING]:
app_list.remove(app)
else:
app['models'].sort(key=lambda x: x['name'])
# Now we order the app list in our defined way in ADMIN_ORDERING (which could be a subset of all apps).
my_ordered_apps = []
if app_dict:
for app_name, object_list in ADMIN_ORDERING:
app = app_dict[app_name]
if object_list == '*':
app['models'].sort(key=lambda x: x['name'])
else:
app['models'].sort(key=lambda x: object_list.index(x['object_name']))
my_ordered_apps.append(app)
# Now we combine and arrange the 2 lists together
my_ordered_apps.extend(app_list)
return my_ordered_apps
# Covering django.contrib.admin.AdminSite.get_app_list
from django.contrib import admin
admin.AdminSite.get_app_list = get_app_list
# =========================================
This is nothing more than overwriting the function defined by Django on the file python2.7/site-packages/django/contrib/admin/sites.py.
That get_app_list method of class AdminSite(object): produces a data structure with all apps on the project, including for Django's auth app, such as:
[
{
"app_label": "auth",
"app_url": "/admin/auth/",
"has_module_perms": "True",
"models": [
{
"add_url": "/admin/auth/group/add/",
"admin_url": "/admin/auth/group/",
"name": "<django.utils.functional.__proxy__ object at 0x11057f990>",
"object_name": "Group",
"perms": {
"add": "True",
"change": "True",
"delete": "True"
}
},
{
"add_url": "/admin/auth/user/add/",
"admin_url": "/admin/auth/user/",
"name": "<django.utils.functional.__proxy__ object at 0x11057f710>",
"object_name": "User",
"perms": {
"add": "True",
"change": "True",
"delete": "True"
}
}
],
"name": "<django.utils.functional.__proxy__ object at 0x108b81850>"
},
{
"app_label": "reservations",
"app_url": "/admin/reservations/",
"has_module_perms": "True",
"models": [
{
"add_url": "/admin/reservations/reservationrule/add/",
"admin_url": "/admin/reservations/reservationrule/",
"name": "<django.utils.functional.__proxy__ object at 0x11057f6d0>",
"object_name": "ReservationRule",
"perms": {
"add": "True",
"change": "True",
"delete": "True"
}
}
],
"name": "Availability"
},
{
"app_label": "blog",
"app_url": "/admin/blog/",
"has_module_perms": "True",
"models": [
{
"add_url": "/admin/blog/category/add/",
"admin_url": "/admin/blog/category/",
"name": "Categories",
"object_name": "Category",
"perms": {
"add": "True",
"change": "True",
"delete": "True"
}
},
{
"add_url": "/admin/blog/post/add/",
"admin_url": "/admin/blog/post/",
"name": "<django.utils.functional.__proxy__ object at 0x11057f110>",
"object_name": "Post",
"perms": {
"add": "True",
"change": "True",
"delete": "True"
}
},
{
"add_url": "/admin/blog/tag/add/",
"admin_url": "/admin/blog/tag/",
"name": "<django.utils.functional.__proxy__ object at 0x11057f390>",
"object_name": "Tag",
"perms": {
"add": "True",
"change": "True",
"delete": "True"
}
}
],
"name": "Blog"
},
(...)
]
This is just a wild stab in the dark, but is there any chance that the order in which you call admin.site.register(< Model class >, < ModelAdmin class >) can determine the display order? Actually, I doubt that would work because I believe Django maintains a registry of the Model -> ModelAdmin objects implemented as a standard Python dictionary, which does not maintain iteration ordering.
If that doesn't behave the way you want, you can always play around with the source in django/contrib/admin. If you need the iteration order maintained, you could replace the _registry object in the AdminSite class (in admin/sites.py) with a UserDict or DictMixin that maintains insertion order for the keys. (But please take this advice with a grain of salt, since I've never made these kinds of changes myself and I'm only making an educated guess at how Django iterates over the collection of ModelAdmin objects. I do think that django/contrib/admin/sites.py is the place to look for this code, though, and the AdminSite class and register() and index() methods in particular are what you want.)
Obviously the nicest thing here would be a simple option for you to specify in your own /admin.py module. I'm sure that's the kind of answer you were hoping to receive. I'm not sure if those options exist, though.
My solution was to make subclasses of django.contrib.admin.sites.AdminSite and django.contrib.admin.options.ModelAdmin .
I did this so I could display a more descriptive title for each app and order the appearance of models in each app. So I have a dict in my settings.py that maps app_labels to descriptive names and the order in which they should appear, the models are ordered by an ordinal field I provide in each ModelAdmin when I register them with the admin site.
Although making your own subclasses of AdminSite and ModelAdmin is encouraged in the docs, my solution looks like an ugly hack in the end.
Copy lib\site-packages\django\contrib\admin\templates\admin\index.html template to project1\templates\admin\ directory, where project1 is the name of your project.
In the copied file, i.e. project1\templates\admin\index.html, replace lines:
{% block content %}
...
{% endblock %}
with:
{% block content %}
<div id="content-main">
{% if app_list %}
<div class="module">
<table>
<caption>App 1</caption>
<tr> <th> Model 1 </th> <td>Description of model 1</td> </tr>
<tr> <th> Model 2 </th> <td>Description of model 1</td> </tr>
<tr> <th> <a href="..." >...</a> </th> <td>...</td> </tr>
</table>
</div>
<div class="module">
<table>
<caption>Authentication and authorization</caption>
<tr> <th> <a href="/admin/auth/user/" >Users</a> </th> <td>List of users</td> </tr>
<tr> <th> <a href="/admin/auth/group/" >Groups</a> </th> <td>List of users' groups</td> </tr>
</table>
</div>
{% else %}
<p>{% trans "You don't have permission to view or edit anything." %}</p>
{% endif %}
</div>
{% endblock %}
where:
app1 is the name of your application with models,
modeli is the name of i-th model in app1.
If you defined more than one application with models in your project, then simply add another table in the above index.html file.
Because we change the template, we can freely change its HTML code. For example we can add a description of the models as it was shown above. You can also restore Add and Change links - I deleted them since I think they are redundant.
The answer is a practical demonstration of the solution from Dave Kasper's answer.
custom_admin.py
from django.contrib.admin import AdminSite
class CustomAdminSite(AdminSite):
def get_urls(self):
urls = super(MyAdminSite, self).get_urls()
return urls
def get_app_list(self, request):
app_dict = self._build_app_dict(request)
ordered_app_list = []
if app_dict:
# TODO: change this dict
admin_ordering = {
'app1': ('Model1', 'Model2'),
'app2': ('Model7', 'Model4'),
'app3': ('Model3',),
}
ordered_app_list = []
for app_key in admin_ordering:
app = app_dict[app_key]
app_ordered_models = []
for model_name in admin_ordering[app_key]:
for model in app_dict[app_key]['models']:
if model['object_name'] == model_name:
app_ordered_models.append(model)
break
app['models'] = app_ordered_models
ordered_app_list.append(app)
return ordered_app_list
admin_site = CustomAdminSite()
urls.py
from custom_admin import admin_site
urlpatterns = [
path('admin/', admin_site.urls),
]
Add the below code to your settings.py:
def get_app_list(self, request):
"""
Return a sorted list of all the installed apps that have been
registered on this site.
"""
ordering = {
# for superuser
'Group': 1,
'User': 2,
# fist app
'TopMessage': 101,
'Slider': 102,
'ProductCategory': 103,
'Product': 104,
'ProductPicture': 105,
# 2nd app
'ProductReview': 201,
'Promotion': 202,
'BestSeller': 203,
}
app_dict = self._build_app_dict(request)
app_list = sorted(app_dict.values(), key=lambda x: x['name'].lower())
for app in app_list:
app['models'].sort(key=lambda x: ordering[x['object_name']])
return app_list
admin.AdminSite.get_app_list = get_app_list
And make changes to the ordering dictionary to match your apps and models. That's it.
The benefit of my solution is that it will show the 'auth' models if the user is a superuser.
Django, by default, orders the models in admin alphabetically. So the order of models in Event admin is Epic, EventHero, EventVillain, Event
Instead you want the order to be
EventHero, EventVillain, Epic then event.
The template used to render the admin index page is admin/index.html and the view function is ModelAdmin.index.
def index(self, request, extra_context=None):
"""
Display the main admin index page, which lists all of the installed
apps that have been registered in this site.
"""
app_list = self.get_app_list(request)
context = {
**self.each_context(request),
'title': self.index_title,
'app_list': app_list,
**(extra_context or {}),
}
request.current_app = self.name
return TemplateResponse(request, self.index_template or
'admin/index.html', context)
The method get_app_list, set the order of the models.:
def get_app_list(self, request):
"""
Return a sorted list of all the installed apps that have been
registered in this site.
"""
app_dict = self._build_app_dict(request)
# Sort the apps alphabetically.
app_list = sorted(app_dict.values(), key=lambda x: x['name'].lower())
# Sort the models alphabetically within each app.
for app in app_list:
app['models'].sort(key=lambda x: x['name'])
return app_list
So to set the order we override get_app_list as:
class EventAdminSite(AdminSite):
def get_app_list(self, request):
"""
Return a sorted list of all the installed apps that have been
registered in this site.
"""
ordering = {
"Event heros": 1,
"Event villains": 2,
"Epics": 3,
"Events": 4
}
app_dict = self._build_app_dict(request)
# a.sort(key=lambda x: b.index(x[0]))
# Sort the apps alphabetically.
app_list = sorted(app_dict.values(), key=lambda x: x['name'].lower())
# Sort the models alphabetically within each app.
for app in app_list:
app['models'].sort(key=lambda x: ordering[x['name']])
return app_list
The code app['models'].sort(key=lambda x: ordering[x['name']]) sets the fixed ordering. Your app now looks like this.
Check the Documentation
In Settings.py file setting the order in which the apps/modules show up listed on Admin
set my ordering list
ADMIN_ORDERING = [
('Your_App1', '*'),
('Your_App2', '*'),
]
Creating a sort function below ADMIN_ORDERING
def get_app_list(self, request):
"""
You Can Set Manually Ordering For Your Apps And Models
ADMIN_ORDERING = [
('Your_App1', [
'module_1',
'module_2'
]),
('Your_App2', '*'),
]
"""
app_dict = self._build_app_dict(request)
# Let's start by sorting the apps alphabetically on a list:
app_list = sorted(app_dict.values(), key=lambda x: x['name'].lower())
# Sorting the models alphabetically within each app.
for app in app_list:
if app['app_label'] in [el[0] for el in ADMIN_ORDERING]:
app_list.remove(app)
else:
app['models'].sort(key=lambda x: x['name'])
# Now we order the app list in our defined way in ADMIN_ORDERING (which could be a subset of all apps).
my_ordered_apps = []
if app_dict:
for app_name, object_list in ADMIN_ORDERING:
app = app_dict[app_name]
if object_list == '*':
app['models'].sort(key=lambda x: x['name'])
else:
app['models'].sort(key=lambda x: object_list.index(x['object_name']))
my_ordered_apps.append(app)
# Now we combine and arrange the 2 lists together
my_ordered_apps.extend(app_list)
return my_ordered_apps
# Covering django.contrib.admin.AdminSite.get_app_list
from django.contrib import admin
admin.AdminSite.get_app_list = get_app_list
Quite a simple and self contained approach can be using the decorator pattern to resort your app and modules in this way:
# admin.py
from django.contrib import admin
def app_resort(func):
def inner(*args, **kwargs):
app_list = func(*args, **kwargs)
# Useful to discover your app and module list:
#import pprint
#pprint.pprint(app_list)
app_sort_key = 'name'
app_ordering = {
"APP_NAME1": 1,
"APP_NAME2": 2,
"APP_NAME3": 3,
}
resorted_app_list = sorted(app_list, key=lambda x: app_ordering[x[app_sort_key]] if x[app_sort_key] in app_ordering else 1000)
model_sort_key = 'object_name'
model_ordering = {
"Model1": 1,
"Model2": 2,
"Model3": 3,
"Model14": 4,
}
for app in resorted_app_list:
app['models'].sort(key=lambda x: model_ordering[x[model_sort_key]] if x[model_sort_key] in model_ordering else 1000)
return resorted_app_list
return inner
admin.site.get_app_list = app_resort(admin.site.get_app_list)
This code sorts at the top only the defined keys in the ordering dict, leaving at the bottom all the other.
The approach is clean but it requires get_app_list to be executed before... which is probably not a good idea when performance is important and you have a large app_list.
This has become a lot easier in Django 4.1:
The AdminSite.get_app_list() method now allows changing the order of apps and models on the admin index page.
You can subclass and override this method to change the order the returned list of apps/models.