我发现了{{请求。Get_host}}方法。

补充Carl Meyer，你可以做一个这样的上下文处理器:

module.context_processors.py

from django.conf import settings

def site(request):
    return {'SITE_URL': settings.SITE_URL}

当地settings.py

SITE_URL = 'http://google.com' # this will reduce the Sites framework db call.

settings.py

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

返回上下文实例的模板，url站点为{{SITE_URL}}

如果想在上下文处理器中处理子域或SSL，可以编写自己的例程。

下面这些可以获得完整的url和部分url:

def myview(request):
    request.build_absolute_uri()
    # http://localhost:8000/admin/store/product/

    request.build_absolute_uri('/')
    # http://localhost:8000/

    request.build_absolute_uri('/')[:-1]
    # http://localhost:8000

    request.scheme
    # http

    request.META['HTTP_HOST']
    # localhost:8000

    request.path    
    # /admin/store/product/

如果你使用“request”上下文处理器，并且正在使用Django sites框架，并且已经安装了Site中间件(即你的设置包括这些):

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

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

．.． 那么您将在模板中拥有可用的请求对象，并且它将包含对请求的当前Site的引用，即request. Site。然后你可以在模板中检索域:

    {{request.site.domain}}

并附上网站名称:

    {{request.site.name}}

我使用自定义模板标记。例如:<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

在模板中使用它，就像这样:

{% load site %}
{% current_domain %}

{{请求。get_host}}在和ALLOWED_HOSTS设置(在Django 1.4.4中添加)一起使用时，可以防止HTTP主机头攻击。

注意{{request.META。HTTP_HOST}}没有相同的保护。查看文档:

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.

至于在模板中使用请求，在Django 1.8中，模板渲染函数调用已经改变了，所以你不再需要直接处理RequestContext。

下面是如何渲染视图的模板，使用快捷函数render():

from django.shortcuts import render

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

下面是如何呈现一个电子邮件模板，IMO是更常见的情况下，你想要的主机值:

from django.template.loader import render_to_string

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

下面是一个在电子邮件模板中添加完整URL的示例;请求。Scheme应该获得HTTP或HTTPS，这取决于您使用的是什么:

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