Changing Site Name in Django 1.9 - python

I am currently following a tutorial which introduces a custom Bootstrap 3 template and builds a Django site using it. In the tutorial they suggest that one changes the following template snippet:
<a class="navbar-brand" href="index.html">Start Bootstrap</a>
to the following snippet.
<a class="navbar-brand" href="/">{{ request.site.name }}</a>
However, when I make this change, no site name shows. I am wondering where I should be setting this name. If it helps, I am using Django CMS and there is only one site called example.compopulated in the Sites section of the Administration.

Had the same problem, found this in the documentation.
If you often use this pattern:
from django.contrib.sites.models import Site
def my_view(request):
site = Site.objects.get_current()
... there is simple way to avoid repetitions. Add django.contrib.sites.middleware.CurrentSiteMiddleware to MIDDLEWARE_CLASSES. The middleware sets the site attribute on every request object, so you can use request.site to get the current site.

You can add site name or change existing site called example.com in the Sites section of the Administration.
To access Site object you need to provide SITE_ID in your settings.py
If you are changing example.com then use SITE_ID = 1 in settings
To render site name in django templates, get value from site model using
from django.contrib.sites.models import Site
current_domain = Site.objects.get_current().domain
then pass current_domain to template

It seems that you have to enable the sites framework, refer to doc.

first yo have to understand that in the setting you have a site_id set to 1 and in database migrations an example site with domain name example.com is created. So, you have two options:
You can change this default from django site models like this: python manage.py shell -c "from django.contrib.sites.models import Site; Site.objects.filter(domain='example.com').update(name='My Site', domain='mysite.tld')" then you restart the server.
You can insert a new row in the site models like this: python manage.py shell -c "from django.contrib.sites.models import Site; mysite,_=Site.objects.get_or_create(id=101, name='mysite.tld', domain='mysite.tld'); print(mysite.id, mysite.name)" Then change the site_id in the settings to the new site_id which should be 2 since it is the second new row.
You can take a look at this Github trend for more clearity.

Related

DJANGO: Set site_id to sites pk

I am following this tutorial for email verification.The version the author is working with is old.I got an error that said
Reverse for ‘activate’ with keyword arguments ‘{‘uidb64’: b’OA’, ‘token’: ‘4tm-3fcfb375c8ba14f9a95b’} . I got that fixed through the first comment . The email got sent .But the link led to www.example.com . The second comment tells how to fix that . The comment was :
For those are using Django 3, You should change some code
Six is deprecated in Django 3, you can use ‘import six’ instead of ‘from django.utils import six’
To send html email, add
email.content_subtype = “html”
after EmailMessage Object.
activate url should be
path(‘activate//’, views.activate, name=’activate’),
get_current_site(request) will return example.com as default when SITE_ID=1 in your settings.py. Add your site name and domail in admin site (/admin/sites/site/) and replace SITE_ID with your sites pk.
But I did not understand how to set SITE_ID to my sites pk.
In settings.py set
BASE_URL = 'https://www.yourdomainname.com'
Don't include trailing /
Also, you have to mention this line in the same file
SITE_ID = 1
Moreover, go to your django admin pannel , default is /admin, and go to the sites tab/model and add/edit your site according to your ip/domainName and display name . This is the important part to do.
If you are testing locally for now then you can use 127.0.0.1:8000 in your ip/domainName and yourdomainNAme.com in display name
If you are deploying or using live then you can use actual ip of your server into ip/domainname or also you can use your domain name (as it's mapped to your ip via DNS system) and display name will be the same.
Hope, it will resolve your problem or issue.

How to get a url map in django? [duplicate]

Is there a way to get the complete django url configuration?
For example Django's debugging 404 page does not show included url configs, so this is not the complete configuration.
Django extensions provides a utility to do this as a manage.py command.
pip install django-extensions
Then add django_extensions to your INSTALLED_APPS in settings.py. then from the console just type the following
python manage.py show_urls
Django is Python, so introspection is your friend.
In the shell, import urls. By looping through urls.urlpatterns, and drilling down through as many layers of included url configurations as possible, you can build the complete url configuration.
import urls
urls.urlpatterns
The list urls.urlpatterns contains RegexURLPattern and RegexURLResolver objects.
For a RegexURLPattern object p you can display the regular expression with
p.regex.pattern
For a RegexURLResolver object q, which represents an included url configuration, you can display the first part of the regular expression with
q.regex.pattern
Then use
q.url_patterns
which will return a further list of RegexURLResolver and RegexURLPattern objects.
At the risk of adding a "me too" answer, I am posting a modified version of the above submitted script that gives you a view listing all the URLs in the project, somewhat prettified and sorted alphabetically, and the views that they call. More of a developer tool than a production page.
def all_urls_view(request):
from your_site.urls import urlpatterns #this import should be inside the function to avoid an import loop
nice_urls = get_urls(urlpatterns) #build the list of urls recursively and then sort it alphabetically
return render(request, "yourapp/links.html", {"links":nice_urls})
def get_urls(raw_urls, nice_urls=[], urlbase=''):
'''Recursively builds a list of all the urls in the current project and the name of their associated view'''
from operator import itemgetter
for entry in raw_urls:
fullurl = (urlbase + entry.regex.pattern).replace('^','')
if entry.callback: #if it points to a view
viewname = entry.callback.func_name
nice_urls.append({"pattern": fullurl,
"location": viewname})
else: #if it points to another urlconf, recur!
get_urls(entry.url_patterns, nice_urls, fullurl)
nice_urls = sorted(nice_urls, key=itemgetter('pattern')) #sort alphabetically
return nice_urls
and the template:
<ul>
{% for link in links %}
<li>
{{link.pattern}} ----- {{link.location}}
</li>
{% endfor%}
</ul>
If you wanted to get real fancy you could render the list with input boxes for any of the regexes that take variables to pass to the view (again as a developer tool rather than production page).
This question is a bit old, but I ran into the same problem and I thought I would discuss my solution. A given Django project obviously needs a means of knowing about all its URLs and needs to be able to do a couple things:
map from a url -> view
map from a named url -> url (then 1 is used to get the view)
map from a view name -> url (then 1 is used to get the view)
Django accomplishes this mostly through an object called a RegexURLResolver.
RegexURLResolver.resolve (map from a url -> view)
RegexURLResolver.reverse
You can get your hands on one of these objects the following way:
from my_proj import urls
from django.core.urlresolvers import get_resolver
resolver = get_resolver(urls)
Then, you can simply print out your urls the following way:
for view, regexes in resolver.reverse_dict.iteritems():
print "%s: %s" % (view, regexes)
That said, Alasdair's solution is perfectly fine and has some advantages, as it prints out some what more nicely than this method. But knowing about and getting your hands on a RegexURLResolver object is something nice to know about, especially if you are interested in Django internals.
The easiest way to get a complete list of registered URLs is to install contrib.admindocs then check the "Views" section. Very easy to set up, and also gives you fully browsable docs on all of your template tags, models, etc.
I have submitted a package (django-showurls) that adds this functionality to any Django project, it's a simple new management command that integrates well with manage.py:
$ python manage.py showurls
^admin/
^$
^login/$
^logout/$
.. etc ..
You can install it through pip:
pip install django-showurls
And then add it to your installed apps in your Django project settings.py file:
INSTALLED_APPS = [
..
'django_showurls',
..
]
And you're ready to go.
More info here -
https://github.com/Niklas9/django-showurls
If you want a list of all the urls in your project, first you need to install django-extensions
You can simply install using command.
pip install django-extensions
For more information related to package goto django-extensions
After that, add django_extensions in INSTALLED_APPS in your settings.py file like this:
INSTALLED_APPS = (
...
'django_extensions',
...
)
urls.py example:
from django.urls import path, include
from . import views
from . import health_views
urlpatterns = [
path('get_url_info', views.get_url_func),
path('health', health_views.service_health_check),
path('service-session/status', views.service_session_status)
]
And then, run any of the command in your terminal
python manage.py show_urls
or
./manage.py show_urls
Sample output example based on config urls.py:
/get_url_info django_app.views.get_url_func
/health django_app.health_views.service_health_check
/service-session/status django_app.views.service_session_status
For more information you can check the documentation.
Are you looking for the urls evaluated or not evaluated as shown in the DEBUG mode? For evaluated, django.contrib.sitemaps can help you there, otherwise it might involve some reverse engineering with Django's code.
When I tried the other answers here, I got this error:
django.core.exceptions.AppRegistryNotReady: Apps aren't loaded yet.
It looks like the problem comes from using django.contrib.admin.autodiscover() in my urls.py, so I can either comment that out, or load Django properly before dumping the URL's. Of course if I want to see the admin URL's in the mapping, I can't comment them out.
The way I found was to create a custom management command that dumps the urls.
# install this file in mysite/myapp/management/commands/urldump.py
from django.core.management.base import BaseCommand
from kive import urls
class Command(BaseCommand):
help = "Dumps all URL's."
def handle(self, *args, **options):
self.show_urls(urls.urlpatterns)
def show_urls(self, urllist, depth=0):
for entry in urllist:
print ' '.join((" " * depth, entry.regex.pattern,
entry.callback and entry.callback.__module__ or '',
entry.callback and entry.callback.func_name or ''))
if hasattr(entry, 'url_patterns'):
self.show_urls(entry.url_patterns, depth + 1)
If you are running Django in debug mode (have DEBUG = True in your settings) and then type a non-existent URL you will get an error page listing the complete URL configuration.

how to run multiple websites and serve different content in django 1.6?

I want to start one project of django and ideally it will have one admin panel .
I want to point different domains and subdomains (subdomains on more priority) and want to serve different content and pages when some1 hit my server .
So to be in details if i have two domains :
1) abc.com
2) xyz.com
Then if someone put abc.com then i should be able to see page1 and all urls associated with this abc.com should be available and should be able to see page2 when someone use xyz.com
and xyz.com/new/ should deliver different content and abc.com/new/ should give different content .
I will suggest you to use Mezzanine for this . It is very strong CMS framework build into django.
To implement multisite application you can use djagno multisite app . It is very easy to integrate .
With the site framework linked into your models, you can associate data to different sites.
You can also use it in your views like in the example taken from the official doc:
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

How does Django build URLs that depend on a site?

I am trying to find out how does Django build URLs, especially those that depend on multiple sites. Where does Django build URLs and how is the site domain prepended to the user defined URL patterns?
Basically, I have several Django CMS pages and multiple Django sites. I need to know how are the URLs created for a page that relies on a different site than the current site; namely, how is the correct site's domain added to a page's URL.
Seems like, adding a Django site's domain to the URL has to be done manually. From Django's docs:
>>> from django.contrib.sites.models import Site
>>> obj = MyModel.objects.get(id=3)
>>> obj.get_absolute_url()
'/mymodel/objects/3/'
>>> Site.objects.get_current().domain
'example.com'
>>> 'http://%s%s' % (Site.objects.get_current().domain, obj.get_absolute_url())
'http://example.com/mymodel/objects/3/'

Django Admin's "view on site" points to example.com instead of my domain

I added a get_absolute_url function to one of my models.
def get_absolute_url(self):
return '/foo/bar'
The admin site picks it up and adds a "view on site" link to the detail page for that object (when I put a real URL there instead of "/foo/bar").
The problem is instead of going to http://localhost:8000/foo/bar, it goes to http://example.com/foo/bar.
What am I doing wrong?
You have to change default site domain value.
The funniest thing is that "example.com" appears in an obvious place. Yet, I was looking for in in an hour or so.
Just use your admin interface -> Sites -> ... there it is :)
You can change this in /admin/sites if you have admin enabled.
As others have mentioned, this is to do with the default sites framework.
If you're using South for database migrations (probably a good
idea in general), you can use a data migration to avoid having to make this same database change everywhere you deploy your application, along the lines of
from south.v2 import DataMigration
from django.conf import settings
class Migration(DataMigration):
def forwards(self, orm):
Site = orm['sites.Site']
site = Site.objects.get(id=settings.SITE_ID)
site.domain = 'yoursite.com'
site.name = 'yoursite'
site.save()
If you are on newer versions of django. the data migration is like this:
from django.conf import settings
from django.db import migrations
def change_site_name(apps, schema_editor):
Site = apps.get_model('sites', 'Site')
site = Site.objects.get(id=settings.SITE_ID)
site.domain = 'yourdomain.com'
site.name = 'Your Site'
site.save()
class Migration(migrations.Migration):
dependencies = [
('app', '0001_initial'),
]
operations = [
migrations.RunPython(change_site_name),
]
When you have edited a Site instance thought the admin, you need to restart your web server for the change to take effect. I guess this must mean that the database is only read when the web server first starts.

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