How do I change a Django app's name in the admin?
I'm using Django 1.6.10
I've tried this:-
class MyModel(models.Model):
pass
class Meta:
app_label = 'My APP name'
but nothing happens.
I strongly recommend you upgrade to the latest 1.7.X, or ideally, Django 1.8.X. Django 1.6 is end of life, and does not support security updates.
The app_label should be a package name, e.g. myapp. It is not meant to be used as a human friendly string e.g. 'My App'. Doing this might break things.
The Django app loading was refactored in Django 1.7. In Django 1.7+, you can change the displayed application name by setting verbose_name for your app config. I'm afraid I don't know of a way to change the display of the app name for Django < 1.7.
write this code at the bottom of urls.py
admin.site.site_header = 'My APP name'
Related
I am trying to work with Sites Model of Django.
I dont quite understand why SITE_ID should be SITE_ID = 1.
in the docs:
The ID, as an integer, of the current site in the django_site database
table. This is used so that application data can hook into specific
sites and a single database can manage content for multiple sites.
why 1? what is the current site? this is not clearly explained in the docs.
lets say, I have www.coolsite.com and some other subdomains like www.wow.coolsite.com and www.awesome.coolsite.com
I want to render different content depending on domain name.
my question is, or better, are:
Do I have to add all those domains into Sites Table in DB?
if so, how should I set SITE_ID in settings? Do I have to set all ids like SITE_ID = 1, SITE_ID = 2.. etc?
what does current site has to do with SITE_ID = 1?
I am a bit confused here.
I thought, each Site (e.g. www.wow.coolsite.com) should be a separate django project so that they can have their own settings.py? and in each of those settings.py's, I will set the id of that page from Sites table? but then there are many django projects which also doesnot make sense to me.
Django was created from a set of scripts developed at a newspaper to publish content on multiple domains; using one single content base.
This is where the "sites" module comes in. Its purpose is to mark content to be displayed for different domains.
In previous versions of django, the startproject script automatically added the django.contrib.sites application to INSTALLED_APPS, and when you did syncdb, a default site with the URL example.com was added to your database, and since this was the first site, its ID was 1 and that's where the setting comes from.
Keep in mind that starting from 1.6, this framework is not enabled by default. So if you need it, you must enable it
The SITE_ID setting sets the default site for your project. So, if you don't specify a site, this is the one it will use.
So to configure your application for different domains:
Enable the sites framework
Change the default site from example.com to whatever your default domain is. You can do this from the django shell, or from the admin.
Add your other sites for which you want to publish content to the sites application. Again, you can do this from the django shell just like any other application or from the admin.
Add a foreign key to the Site model in your object site = models.ForeignKey(Site)
Add the site manager on_site = CurrentSiteManager()
Now, when you want to filter content for the default site, or a particular site:
foo = MyObj.on_site.all() # Filters site to whatever is `SITE_ID`
foo = MyObj.objects.all() # Get all objects, irrespective of what site
# they belong to
The documentation has a full set of examples.
Things would be much easier to understand if Django's default SiteAdmin included the id field in the list_display fields.
To do this, you can redefine SiteAdmin (anywhere in your app, but I'd recommend your admin.py or maybe your urls.py) like this:
from django.contrib import admin
from django.contrib.sites.models import Site
admin.site.unregister(Site)
class SiteAdmin(admin.ModelAdmin):
fields = ('id', 'name', 'domain')
readonly_fields = ('id',)
list_display = ('id', 'name', 'domain')
list_display_links = ('name',)
search_fields = ('name', 'domain')
admin.site.register(Site, SiteAdmin)
After including this code snippet, the ID for each "Site" will be shown in the first column of the admin list and inside the form as a read only field. These 'id' fields are what you need to use as SITE_ID:
The concept is that each different site runs in a different application server instance, launched using a different yourdomain_settings.py that then includes a base_settings.py with the rest of the common configuration.
Each of these yourdomain_settings.py will define its own SITE_ID and all other different settings.py parameters that they need to look and be different from each other (static resources, templates, etc.) then you'll define a DJANGO_SETTINGS_MODULE environment variable pointing to that specific yourdomain_settings.py file when launching the application server instance for that domain.
A further note: get_current_site(request) does need request to be available for it to work. If your code doesn't have one, you can use Site.objects.get_current() that however will need a SITE_ID properly defined in the running application server's settings.
This is a late answer but for anyone else having SITE_ID issues and Site problems.
Inside the database, django has a django_site table with(id, domain, name). This is where django stores the SITE_IDs. Mine was actually 5 in the database but i had it set to SITE_ID=1 in the settings.
Knowing that, i can now go back to the database and clear it to get back to zero or use the actual id in the database.
This is covered in the documentation for the Sites framework:
In order to serve different sites in production, you’d create a
separate settings file with each SITE_ID (perhaps importing from a
common settings file to avoid duplicating shared settings) and then
specify the appropriate DJANGO_SETTINGS_MODULE for each site.
But if you didn't want to do it that way, you can not set SITE_ID at all and just look up the current site based on the domain name in your views using get_current_site:
from django.contrib.sites.shortcuts import get_current_site
def my_view(request):
current_site = get_current_site(request)
if current_site.domain == 'foo.com':
# Do something
pass
else:
# Do something else.
pass
This link explains it:
You’ll want to create separate settings files for each domain you’re adding; each one will need its own MEDIA_URL and other settings. You’ll also want to do two things to make sure everything works out properly for administering the different sites:
Create a new Site object in your admin for each domain, and put the id of that Site into its settings file as SITE_ID so Django knows which site in the database corresponds to this settings file.
In the settings file for your original site (the one with id 1), add the other sites’ settings files to the ADMIN_FOR setting, to let Django know that this one instance of the admin application will handle all of the sites.
Also, if you wanna figure out how to modify models and set the views You may take a look at this link:
https://django.cowhite.com/blog/managing-multiple-websites-with-a-common-database-in-django-the-sites-framework/
I'm a front-end dev struggling along with Django. I have the basics pretty much down but I've hit at wall at the following point.
I have a site running locally and also on a dev machine. Locally I've added an extra class model to an already existing app, registered it in the relevant admin.py and checked it in the settings. Locally the new class and relevant fields appear in admin but when I move this all to dev they're not appearing. The app is called 'publish'.
My method was as follows:
Created the new class in the publish > models.py file:
class Whitepaper(models.Model):
title = models.CharField(max_length=200)
slug = models.SlugField(max_length=100, blank=True)
pub_date = models.DateField('date published')
section = models.ForeignKey('Section', related_name='whitepapers', blank=True, null=True)
description = models.CharField(max_length=1000)
docfile = models.FileField(upload_to="whitepapers/%Y/%m/%d", null=True, blank=True)
Updated and migrated the model with South using:
python manage.py schemamigration publish --auto
and
python manage.py migrate publish
Registered the class in the admin.py file:
from models import Section, Tag, Post, Whitepaper
from django.contrib import admin
from django import forms
admin.site.register(Whitepaper)
The app is listed in the settings.py file:
INSTALLED_APPS = (
...,
...,
'publish',
...,
)
As this is running on a dev server that's hosting a few other testing areas, restarting the whole thing is out of the question so I've been 'touching' the .wsgi file.
On my local version this got the model and fields showing up in the admin but on the dev server they are nowhere to be seen.
What am I missing?
Thanks ye brainy ones.
I figured out the problem. Turns out the login I was using to get into the admin didn't have superuser privileges. So I made a new one with:
python manage.py createsuperuser
After logging in with the new username and password I could see all my new shiny tables!
Are you sure touching .wsgi file does restart your app?
It looks like it doesn't.
Make sure the app is restarted. Find the evidence touching .wsgi file restarts the app maybe.
Since you don't provide any insight about how the dev server runs the apps, we won't be able to help you any further.
I have extended the User model, but now the new user model is in an app called "account" which gives all models inside this app the app label "account". The Django model "Groups" still has the app label "Auth", so now models which all has something to do with auth is in separate apps in the admin site. Is it possibly to change the app label for "Groups"?
Try this:
from django.db.models.loading import get_models
get_models(django.contrib.auth.models)[1]._meta.app_label = 'group' #or whatever
If you need still more flexibility in the admin you could try django-admin-tools. It makes it easy to reorder and group models in different layouts (tabs, collapsible boxes, etc.) and also add dashboard-like features.
Just in case anyone needs this in Django 3.0+:
from django.apps import apps
apps.get_model('auth', 'Group')._meta.app_label = 'group' #or whatever, but have to be a registered app
Please not that this will mess with django internal model handling, e.g. generate migrations in contrib.auth sitepackage and so on
I have a Django project for a simple blog/forum website I’m building.
I’m using the syndication feed framework, which seems to generate the URLs for items in the feed using the domain of the current site from the Sites framework.
I was previously unaware of the Sites framework. My project isn’t going to be used for multiple sites, just one.
What I want to do is set the domain property of the current site. Where in my Django project should I do that? Somewhere in /settings.py?
If I understand correctly, Sites framework data is stored in the database, so if I want to store this permanently, I guess it’s appropriate in an initial_data fixture.
I fired up the Django shell, and did the following:
>>> from django.contrib.sites.models import Site
>>> one = Site.objects.all()[0]
>>> one.domain = 'myveryspecialdomain.com'
>>> one.name = 'My Special Site Name'
>>> one.save()
I then grabbed just this data at the command line:
python manage.py dumpdata sites
And pasted it into my pre-existing initial_data fixture.
The other answers suggest to manually update the site in the admin, shell, or your DB. That's a bad idea—it should be automatic.
You can create a migration that'll do this automatically when you run your migrations, so you can be assured it's always applied (such as when you deploy to production). This is also recommended in the documentation, but it doesn't list instructions.
First, run ./manage.py makemigrations --empty --name UPDATE_SITE_NAME myapp to create an empty migration. Then add the following code:
from django.db import migrations
from django.conf import settings
def update_site_name(apps, schema_editor):
SiteModel = apps.get_model('sites', 'Site')
domain = 'mydomain.com'
SiteModel.objects.update_or_create(
pk=settings.SITE_ID,
defaults={'domain': domain,
'name': domain}
)
class Migration(migrations.Migration):
dependencies = [
# Make sure the dependency that was here by default is also included here
('sites', '0002_alter_domain_unique'), # Required to reference `sites` in `apps.get_model()`
]
operations = [
migrations.RunPython(update_site_name),
]
Make sure you've set SITE_ID in your settings. Then run ./manage.py migrate to apply the changes :)
You can change it using django admin site.
Just go to 127.0.0.1:8000/admin/sites/
For those who are struggling to find "Sites" section on Django's admin page, for newer Django versions you need to enable the optional Sites Framework like so:
On your settings.py file, add this to your "INSTALLED_APPS":
'django.contrib.sites'
Then specify an ID for the default site (since it's probably your first site to be specified, you can use ID 1):
SITE_ID = 1
Run your migrations and check if the "Sites" section is available on your Django's admin page.
More details at https://docs.djangoproject.com/en/3.0/ref/contrib/sites/#enabling-the-sites-framework
You can modify the Site entry in your database manually. Navigate to the table called 'django_site'. Then, you should only see one entry (row). You'll want to modify the field (column) named 'domain'.
I'm hosting a Django site through mod_wsgi and WAMP with Python 2.7. On my admin site the Users, Groups and Sites sections all have Add and Change buttons. While there is a section each for both of my own custom models, Feedpost and Newspost, they have no buttons at all and there is no way to add or change them.
I believe this change came about when I switched from using the Django internal testing server to using WAMP. Does anyone know what's causing this and how to get my Add and Change buttons back?
EDIT:
Here's one of the two models:
from django.db import models
from django.contrib import admin
class FeedPost(models.Model):
title = models.CharField(max_length=50)
text = models.CharField(max_length=5000)
date = models.DateTimeField('Date Published')
def __unicode__(self):
return self.title
admin.site.register(FeedPost)
The other is nearly exactly the same.
It turns out that using the Django development server localhost:8000/admin insists on going to my site's homepage and I can't access the admin site at all.
Also I think this is going to turn out to be the problem because I haven't got an admin.py file anywhere in my project...
These need to be in the admin.py file for starters.
from django.contrib import admin
admin.site.register(FeedPost)