How to get the domain name of my site within a Django template?
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Track title: CC P Beethoven - Piano Sonata No 2 in A
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Chapters
00:00 Question
00:23 Accepted answer (Score 90)
00:37 Answer 2 (Score 135)
01:24 Answer 3 (Score 122)
01:38 Answer 4 (Score 68)
02:27 Thank you
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Full question
https://stackoverflow.com/questions/1451...
Answer 1 links:
[contrib.sites framework]: http://docs.djangoproject.com/en/dev/ref...
Answer 2 links:
[{{ request.get_host }}]: https://docs.djangoproject.com/en/dev/re...
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Tags
#python #python3x #django #djangotemplates #djangotemplatefilters
#avk47
ANSWER 1
Score 139
If you want the actual HTTP Host header, see Daniel Roseman's comment on @Phsiao's answer. The other alternative is if you're using the contrib.sites framework, you can set a canonical domain name for a Site in the database (mapping the request domain to a settings file with the proper SITE_ID is something you have to do yourself via your webserver setup). In that case you're looking for:
from django.contrib.sites.models import Site
current_site = Site.objects.get_current()
current_site.domain
you'd have to put the current_site object into a template context yourself if you want to use it. If you're using it all over the place, you could package that up in a template context processor.
ANSWER 2
Score 137
I've discovered the {{ request.get_host }} method.
ACCEPTED ANSWER
Score 91
I think what you want is to have access to the request context, see RequestContext.
ANSWER 4
Score 71
Complementing Carl Meyer, you can make a context processor like this:
module.context_processors.py
from django.conf import settings
def site(request):
return {'SITE_URL': settings.SITE_URL}
local settings.py
SITE_URL = 'http://google.com' # this will reduce the Sites framework db call.
settings.py
TEMPLATE_CONTEXT_PROCESSORS = (
...
"module.context_processors.site",
....
)
templates returning context instance the url site is {{ SITE_URL }}
you can write your own rutine if want to handle subdomains or SSL in the context processor.