I would like to provide the same content inside 2 different base files.

So I’m trying to do this:

page1.html:

{% extends "base1.html" %}
{% include "commondata.html" %}

page2.html:

{% extends "base2.html" %} 
{% include "commondata.html" %}

The problem is that I can’t seem to use both extends and include. Is there some way to do that? And if not, how can I accomplish the above?

commondata.html overrides a block that is specified in both base1.html and base2.html

The purpose of this is to provide the same page in both pdf and html format, where the formatting is slightly different. The above question though simplifies what I’m trying to do so if I can get an answer to that it will solve my problem.

When you use the extends template tag, you’re saying that the current template extends another — that it is a child template, dependent on a parent template. Django will look at your child template and use its content to populate the parent.

Everything that you want to use in a child template should be within blocks, which Django uses to populate the parent. If you want use an include statement in that child template, you have to put it within a block, for Django to make sense of it. Otherwise it just doesn’t make sense and Django doesn’t know what to do with it.

The Django documentation has a few really good examples of using blocks to replace blocks in the parent template.

https://docs.djangoproject.com/en/dev/ref/templates/language/#template-inheritance

From Django docs:

The include tag should be considered as an implementation of “render this subtemplate and include the HTML”, not as “parse this subtemplate and include its contents as if it were part of the parent”. This means that there is no shared state between included templates — each include is a completely independent rendering process.

So Django doesn’t grab any blocks from your commondata.html and it doesn’t know what to do with rendered html outside blocks.

This should do the trick for you: put include tag inside of a block section.

page1.html:

{% extends "base1.html" %}

{% block foo %}
   {% include "commondata.html" %}
{% endblock %}

page2.html:

{% extends "base2.html" %}

{% block bar %}
   {% include "commondata.html" %}
{% endblock %}

More info about why it wasn’t working for me in case it helps future people:

The reason why it wasn’t working is that {% include %} in django doesn’t like special characters like fancy apostrophe. The template data I was trying to include was pasted from word. I had to manually remove all of these special characters and then it included successfully.

You can’t pull in blocks from an included file into a child template to override the parent template’s blocks. However, you can specify a parent in a variable and have the base template specified in the context.

From the documentation:

{% extends variable %} uses the value of variable. If the variable evaluates to a string, Django will use that string as the name of the parent template. If the variable evaluates to a Template object, Django will use that object as the parent template.

Instead of separate “page1.html” and “page2.html”, put {% extends base_template %} at the top of “commondata.html”. And then in your view, define base_template to be either “base1.html” or “base2.html”.

Added for reference to future people who find this via google: You might want to look at the {% overextend %} tag provided by the mezzanine library for cases like this.

Edit 10th Dec 2015: As pointed out in the comments, ssi is deprecated since version 1.8. According to the documentation:

This tag has been deprecated and will be removed in Django 1.10. Use the include tag instead.

In my opinion, the right (best) answer to this question is the one from podshumok, as it explains why the behaviour of include when used along with inheritance.

However, I was somewhat surprised that nobody mentioned the ssi tag provided by the Django templating system, which is specifically designed for inline including an external piece of text. Here, inline means the external text will not be interpreted, parsed or interpolated, but simply “copied” inside the calling template.

Please, refer to the documentation for further details (be sure to check your appropriate version of Django in the selector at the lower right part of the page).

https://docs.djangoproject.com/en/dev/ref/templates/builtins/#ssi

From the documentation:

ssi
Outputs the contents of a given file into the page.
Like a simple include tag, {% ssi %} includes the contents of another file
– which must be specified using an absolute path – in the current page

Beware also of the security implications of this technique and also of the required ALLOWED_INCLUDE_ROOTS define, which must be added to your settings files.

This may be simple, but I looked around and couldn’t find an answer. What’s the best way to reference a single item in a list from a Django template?

In other words how do I do the equivalent of {{ data[0] }} within the template language?

Thanks.

It looks like {{ data.0 }}. See Variables and lookups.

A better way: custom template filter: https://docs.djangoproject.com/en/dev/howto/custom-template-tags/

such as get my_list[x] in templates:

in template

{% load index %}
{{ my_list|index:x }}

templatetags/index.py

from django import template
register = template.Library()

@register.filter
def index(indexable, i):
    return indexable[i]

if my_list = [['a','b','c'], ['d','e','f']], you can use {{ my_list|index:x|index:y }} in template to get my_list[x][y]

It works fine with “for”

{{ my_list|index:forloop.counter0 }}

Tested and works well ^_^

{{ data.0 }} should work.

Let’s say you wrote data.obj django tries data.obj and data.obj(). If they don’t work it tries data["obj"]. In your case data[0] can be written as {{ data.0 }}. But I recommend you to pull data[0] in the view and send it as separate variable.

@jennifer06262016, you can definitely add another filter to return the objects inside a django Queryset.

@register.filter 
def get_item(Queryset):
    return Queryset.your_item_key

In that case, you would type something like this {{ Queryset|index:x|get_item }} into your template to access some dictionary object. It works for me.

I want to do this:

100 - {{ object.article.rating_score }} 

So for example, the output would be 20 if {{ object.article.rating_score }} equaled 80.

How do I do this at the template level? I don’t have access to the Python code.

You can use the add filter:

{{ object.article.rating_score|add:"-100" }}

Use django-mathfilters. In addition to the built-in add filter, it provides filters to subtract, multiply, divide, and take the absolute value.

For the specific example above, you would use {{ 100|sub:object.article.rating_score }}.

Generally it is recommended you do this calculation in your view. Otherwise, you could use the add filter.

How do I get the domain name of my current site from within a Django template? I’ve tried looking in the tag and filters but nothing there.

I think what you want is to have access to the request context, see RequestContext.

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.

I’ve discovered the {{ request.get_host }} method.

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.

The variation of the context processor I use is:

from django.contrib.sites.shortcuts import get_current_site
from django.utils.functional import SimpleLazyObject


def site(request):
    return {
        'site': SimpleLazyObject(lambda: get_current_site(request)),
    }

The SimpleLazyObject wrapper makes sure the DB call only happens when the template actually uses the site object. This removes the query from the admin pages. It also caches the result.

and include it in the settings:

TEMPLATE_CONTEXT_PROCESSORS = (
    ...
    "module.context_processors.site",
    ....
)

In the template, you can use {{ site.domain }} to get the current domain name.

edit: to support protocol switching too, use:

def site(request):
    site = SimpleLazyObject(lambda: get_current_site(request))
    protocol = 'https' if request.is_secure() else 'http'

    return {
        'site': site,
        'site_root': SimpleLazyObject(lambda: "{0}://{1}".format(protocol, site.domain)),
    }

I know this question is old, but I stumbled upon it looking for a pythonic way to get current domain.

def myview(request):
    domain = request.build_absolute_uri('/')[:-1]
    # that will build the complete domain: http://foobar.com

Quick and simple, but not good for production:

(in a view)

    request.scheme               # http or https
    request.META['HTTP_HOST']    # example.com
    request.path                 # /some/content/1/

(in a template)

{{ request.scheme }} :// {{ request.META.HTTP_HOST }} {{ request.path }}

Be sure to use a RequestContext, which is the case if you’re using render.

Don’t trust request.META['HTTP_HOST'] in production: that info comes from the browser. Instead, use @CarlMeyer’s answer

{{ request.get_host }} should protect against HTTP Host header attacks when used together with the ALLOWED_HOSTS setting (added in Django 1.4.4).

Note that {{ request.META.HTTP_HOST }} does not have the same protection. See the docs:

ALLOWED_HOSTS

A list of strings representing the host/domain names that this Django site can serve. This is a security measure to prevent HTTP Host header attacks, which are possible even under many seemingly-safe web server configurations.

… If the Host header (or X-Forwarded-Host if USE_X_FORWARDED_HOST is enabled) does not match any value in this list, the django.http.HttpRequest.get_host() method will raise SuspiciousOperation.

… This validation only applies via get_host(); if your code accesses the Host header directly from request.META you are bypassing this security protection.

As for using the request in your template, the template-rendering function calls have changed in Django 1.8, so you no longer have to handle RequestContext directly.

Here’s how to render a template for a view, using the shortcut function render():

from django.shortcuts import render

def my_view(request):
    ...
    return render(request, 'my_template.html', context)

Here’s how to render a template for an email, which IMO is the more common case where you’d want the host value:

from django.template.loader import render_to_string

def my_view(request):
    ...
    email_body = render_to_string(
        'my_template.txt', context, request=request)

Here’s an example of adding a full URL in an email template; request.scheme should get http or https depending on what you’re using:

Thanks for registering! Here's your activation link:
{{ request.scheme }}://{{ request.get_host }}{% url 'registration_activate' activation_key %}

I use a custom template tag. Add to e.g. <your_app>/templatetags/site.py:

# -*- coding: utf-8 -*-
from django import template
from django.contrib.sites.models import Site

register = template.Library()

@register.simple_tag
def current_domain():
    return 'http://%s' % Site.objects.get_current().domain

Use it in a template like this:

{% load site %}
{% current_domain %}

Similar to user panchicore’s reply, this is what I did on a very simple website. It provides a few variables and makes them available on the template.

SITE_URL would hold a value like example.com
SITE_PROTOCOL would hold a value like http or https
SITE_PROTOCOL_URL would hold a value like http://example.com or https://example.com
SITE_PROTOCOL_RELATIVE_URL would hold a value like //example.com.

module/context_processors.py

from django.conf import settings

def site(request):

    SITE_PROTOCOL_RELATIVE_URL = '//' + settings.SITE_URL

    SITE_PROTOCOL = 'http'
    if request.is_secure():
        SITE_PROTOCOL = 'https'

    SITE_PROTOCOL_URL = SITE_PROTOCOL + '://' + settings.SITE_URL

    return {
        'SITE_URL': settings.SITE_URL,
        'SITE_PROTOCOL': SITE_PROTOCOL,
        'SITE_PROTOCOL_URL': SITE_PROTOCOL_URL,
        'SITE_PROTOCOL_RELATIVE_URL': SITE_PROTOCOL_RELATIVE_URL
    }

settings.py

TEMPLATE_CONTEXT_PROCESSORS = (
    ...
    "module.context_processors.site",
    ....
 )

SITE_URL = 'example.com'

Then, on your templates, use them as {{ SITE_URL }}, {{ SITE_PROTOCOL }}, {{ SITE_PROTOCOL_URL }} and {{ SITE_PROTOCOL_RELATIVE_URL }}

In a Django template you can do:

<a href="{{ request.scheme }}://{{ request.META.HTTP_HOST }}{{ request.path }}?{{ request.GET.urlencode }}" >link</a>

If you use the “request” context processor, and are using the Django sites framework, and have the Site middleware installed (i.e. your settings include these):

INSTALLED_APPS = [
    ...
    "django.contrib.sites",
    ...
]

MIDDLEWARE = [
    ...
     "django.contrib.sites.middleware.CurrentSiteMiddleware",
    ...
]

… then you will have the request object available in templates, and it will contain a reference to the current Site for the request as request.site. You can then retrieve the domain in a template with:

    {{request.site.domain}}

and the site name with:

    {{request.site.name}}

What about this approach? Works for me. It is also used in django-registration.

def get_request_root_url(self):
    scheme = 'https' if self.request.is_secure() else 'http'
    site = get_current_site(self.request)
    return '%s://%s' % (scheme, site)

from django.contrib.sites.models import Site
if Site._meta.installed:
    site = Site.objects.get_current()
else:
    site = RequestSite(request)

You can use {{ protocol }}://{{ domain }} in your templates to get your domain name.