blocks|key|1449147|text|我认为您想要的是能够访问请求上下文，请参见RequestContext。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449148|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

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

blocks|key|1234551|text|如果您想要实际的HTTP+Host标头，请参阅Daniel+Roseman对@Phsiao的回答的评论。另一种选择是，如果您使用的是contrib.sites+framework，您可以在数据库中为站点设置一个规范的域名(将请求域映射到具有适当SITE_ID的设置文件是您必须通过您的set服务器设置自己完成的事情)。在这种情况下，您正在寻找：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1234552|from+django.contrib.sites.models+import+Site

current_site+=+Site.objects.get_current()
current_site.domain|code-block|syntax|javascript|1234553|如果您想要使用current_site对象，则必须自己将其放入模板上下文中。如果你到处都在使用它，你可以把它打包到一个模板上下文处理器中。|1234554|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.djangoproject.com/en/dev/ref/contrib/sites/#ref-contrib-sites^0|1U|N|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

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 <a href="http://docs.djangoproject.com/en/dev/ref/contrib/sites/#ref-contrib-sites" rel="noreferrer">contrib.sites framework</a>, 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:

<pre><code>from django.contrib.sites.models import Site

current_site = Site.objects.get_current()
current_site.domain
</code></pre>

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.

blocks|key|1449300|text|作为对Carl+Meyer的补充，您可以像这样制作一个上下文处理器：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449301|module.context_processors.py|1449302|from+django.conf+import+settings

def+site(request):
++++return+{'SITE_URL':+settings.SITE_URL}|code-block|syntax|javascript|1449303|本地settings.py|1449304|SITE_URL+=+'http://google.com'+#+this+will+reduce+the+Sites+framework+db+call.|1449305|settings.py|1449306|TEMPLATE_CONTEXT_PROCESSORS+=+(
++++...
++++"module.context_processors.site",
++++....
+)|1449307|返回上下文实例的模板url站点为{{+SITE_URL+}}|1449308|如果想要在上下文处理器中处理子域或SSL，您可以编写自己的规则。|1449309|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Complementing Carl Meyer, you can make a context processor like this:

<h3>module.context_processors.py</h3>

<pre><code>from django.conf import settings

def site(request):
 return {'SITE_URL': settings.SITE_URL}
</code></pre>

<h3>local settings.py</h3>

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

<h3>settings.py</h3>

<pre><code>TEMPLATE_CONTEXT_PROCESSORS = (
 ...
 "module.context_processors.site",
 ....
 )
</code></pre>

<h3>templates returning context instance the url site is {{ SITE_URL }}</h3>

you can write your own rutine if want to handle subdomains or SSL in the context processor.

blocks|key|1449360|text|我发现了{{+request.get_host+}}方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1449361|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/dev/ref/request-response/#django.http.HttpRequest.get_host^0|4|M|4|M|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

I've discovered the <a href="https://docs.djangoproject.com/en/dev/ref/request-response/#django.http.HttpRequest.get_host" rel="noreferrer"><code>{{ request.get_host }}</code></a> method.

blocks|key|1449419|text|您可以在模板中使用{{+protocol+}}://{{+domain+}}来获取您的域名。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1449420|entityMap^0|9|T|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

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

blocks|key|1449455|text|我使用一个定制的模板标签。添加到例如<your_app>/templatetags/site.py|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1449456|#+-*-+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://%25s'+%25+Site.objects.get_current().domain|code-block|syntax|javascript|1449457|在模板中使用它，如下所示：|1449458|{%25+load+site+%25}
{%25+current_domain+%25}|1449459|entityMap^0|I|V|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

I use a custom template tag. Add to e.g. <code>&lt;your_app&gt;/templatetags/site.py</code>:

<pre><code># -*- 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
</code></pre>

Use it in a template like this:

<pre><code>{% load site %}
{% current_domain %}
</code></pre>

blocks|key|1449470|text|from+django.contrib.sites.models+import+Site
if+Site._meta.installed:
++++site+=+Site.objects.get_current()
else:
++++site+=+RequestSite(request)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1449471|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

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

blocks|key|1234823|text|快速简单，但不利于生产：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1234824|(在视图中)|1234825|++++request.scheme+++++++++++++++#+http+or+https
++++request.META['HTTP_HOST']++++#+example.com
++++request.path+++++++++++++++++#+/some/content/1/|code-block|syntax|javascript|1234826|(在模板中)|1234827|{{+request.scheme+}}+://+{{+request.META.HTTP_HOST+}}+{{+request.path+}}|1234828|请确保使用RequestContext，如果您使用的是render，就是这种情况。|offset|length|1234829|在生产环境中不要相信request.META['HTTP_HOST']：这些信息来自浏览器。相反，使用@CarlMeyer的答案|style|CODE|1234830|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/1.7/ref/templates/api/#django.template.RequestContext|1|https://docs.djangoproject.com/en/1.7/topics/http/shortcuts/#render^0|0|0|0|0|0|5|E|0|R|6|1|0|A|P|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|18|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|19|8|@]|9|@$O|1A|P|1B|1|1C]|$O|1D|P|1E|1|1F]]|A|$]]|$1|Q|3|R|5|6|7|1G|8|@$O|1H|P|1I|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]|12|$5|X|Y|Z|A|$10|13]]]]

Quick and simple, but not good for production:

(in a view)

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

(in a template)

<pre><code>{{ request.scheme }} :// {{ request.META.HTTP_HOST }} {{ request.path }}
</code></pre>

Be sure to use a <a href="https://docs.djangoproject.com/en/1.7/ref/templates/api/#django.template.RequestContext">RequestContext</a>, which is the case if you're using <a href="https://docs.djangoproject.com/en/1.7/topics/http/shortcuts/#render">render</a>. 

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

blocks|key|1234874|text|我知道这个问题很老了，但我偶然发现了这个问题，为了寻找一种获取当前域名的pythonic方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1234875|def+myview(request):
++++domain+=+request.build_absolute_uri('/')[:-1]
++++#+that+will+build+the+complete+domain:+http://foobar.com|code-block|syntax|javascript|1234876|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

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

blocks|key|1449645|text|这种方法怎么样？这对我很有效。它也用在django-registration中。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1449646|def+get_request_root_url(self):
++++scheme+=+'https'+if+self.request.is_secure()+else+'http'
++++site+=+get_current_site(self.request)
++++return+'%25s://%25s'+%25+(scheme,+site)|code-block|syntax|javascript|1449647|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/ubernostrum/django-registration/blob/8c0acebbec298ce8971650b93abcbc07d0326354/registration/backends/hmac/views.py#L74^0|J|J|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

What about this approach? Works for me.
It is also used in <a href="https://github.com/ubernostrum/django-registration/blob/8c0acebbec298ce8971650b93abcbc07d0326354/registration/backends/hmac/views.py#L74" rel="nofollow noreferrer">django-registration</a>.

<pre><code>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)
</code></pre>

blocks|key|1449697|text|在Django模板中，您可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449698|<a+href="{{+request.scheme+}}://{{+request.META.HTTP_HOST+}}{{+request.path+}}?{{+request.GET.urlencode+}}"+>link</a>|code-block|syntax|javascript|1449699|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In a Django template you can do:

<pre><code>&lt;a href="{{ request.scheme }}://{{ request.META.HTTP_HOST }}{{ request.path }}?{{ request.GET.urlencode }}" &gt;link&lt;/a&gt;
</code></pre>

blocks|key|1235005|text|如果您使用"request"+context+processor，并且正在使用Django+sites+framework，并且安装了Site+middleware+(即，您的设置包括以下内容)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1235006|INSTALLED_APPS+=+[
++++...
++++"django.contrib.sites",
++++...
]

MIDDLEWARE+=+[
++++...
+++++"django.contrib.sites.middleware.CurrentSiteMiddleware",
++++...
]|code-block|syntax|javascript|1235007|..。然后，您将在模板中获得request对象，并且它将包含对请求的当前Site的引用，作为request.site。然后，您可以使用以下命令在模板中检索该域：|style|CODE|1235008|++++{{request.site.domain}}|1235009|和站点名称，使用：|1235010|++++{{request.site.name}}|1235011|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/3.1/ref/templates/api/#django-template-context-processors-request|1|https://docs.djangoproject.com/en/3.1/ref/contrib/sites/|2|https://docs.djangoproject.com/en/3.1/ref/contrib/sites/#site-middleware^0|5|R|0|13|M|1|1V|F|2|0|0|E|7|10|4|1A|C|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]|$A|18|B|19|1|1A]|$A|1B|B|1C|1|1D]]|C|$]]|$1|D|3|E|5|F|7|1E|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1F|8|@$A|1G|B|1H|K|L]|$A|1I|B|1J|K|L]|$A|1K|B|1L|K|L]]|9|@]|C|$]]|$1|M|3|N|5|F|7|1M|8|@]|9|@]|C|$G|H]]|$1|O|3|P|5|6|7|1N|8|@]|9|@]|C|$]]|$1|Q|3|R|5|F|7|1O|8|@]|9|@]|C|$G|H]]|$1|S|3|-4|5|6|7|1P|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]|10|$5|V|W|X|C|$Y|11]]|12|$5|V|W|X|C|$Y|13]]]]

If you use the <a href="https://docs.djangoproject.com/en/3.1/ref/templates/api/#django-template-context-processors-request" rel="noreferrer">&quot;request&quot; context processor</a>, and are using the <a href="https://docs.djangoproject.com/en/3.1/ref/contrib/sites/" rel="noreferrer">Django sites framework</a>, and have the <a href="https://docs.djangoproject.com/en/3.1/ref/contrib/sites/#site-middleware" rel="noreferrer">Site middleware</a> installed (i.e. your settings include these):
<pre class="lang-py prettyprint-override"><code>INSTALLED_APPS = [
 ...
 &quot;django.contrib.sites&quot;,
 ...
]

MIDDLEWARE = [
 ...
 &quot;django.contrib.sites.middleware.CurrentSiteMiddleware&quot;,
 ...
]
</code></pre>
... then you will have the <code>request</code> object available in templates, and it will contain a reference to the current <code>Site</code> for the request as <code>request.site</code>. You can then retrieve the domain in a template with:
<pre><code> {{request.site.domain}}
</code></pre>
and the site name with:
<pre><code> {{request.site.name}}
</code></pre>

blocks|key|1449778|text|您可以使用request.build_absolute_uri()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1449779|默认情况下，它将返回完整路径。|1449780|但是如果你像这样传入一个参数：|1449781|request.build_absolute_uri('/')|code-block|syntax|javascript|1449782|这将返回域名。|1449783|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/3.1/ref/request-response/#django.http.HttpRequest.build_absolute_uri^0|5|S|5|S|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|6|7|14|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|15|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|16|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|17|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

You can use <a href="https://docs.djangoproject.com/en/3.1/ref/request-response/#django.http.HttpRequest.build_absolute_uri" rel="nofollow noreferrer"><code>request.build_absolute_uri()</code></a>
By default, it will return a full path.
But if you pass in a parameter like this:
<pre><code>request.build_absolute_uri('/')
</code></pre>
This would return the domain name.

blocks|key|1449850|text|我认为我们想要的是现有url标签的替代，所以我写了一个新标签：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1449851|from+django.template+import+Library
from+django.urls+++++import+reverse

@register.simple_tag(takes_context+=+True)
def+fullURL(context,+name,+*args,+**kwargs):
++++request+=+context['request']
++++return+f'{request.scheme}://{request.get_host()}{reverse(name,+args+=+args,+kwargs+=+kwargs)}'|code-block|syntax|javascript|1449852|然后在你的模板中你可以有这个。|1449853|{%25+extends+"myapp/email/email_base.html"+%25}

{%25+load+mytags+%25}+{#+Replace+mytags+with+whatever+the+name+of+your+custom+tags+calss+is.+#}

{%25+block+content+%25}
You+can+use+<a+href="{%25+fullURL+'signup'+%25}">this+link</a>+to+get+started+with+your+account.+We+look+forward+to+seeing+you+soon!
{%25+endblock+content+%25}|1449854|然后，当您生成此代码时，您只需记住将request传递给context，如下所示……|1449855|from+django.template.loader+import+render_to_string

def+sendEmail(subject,+to,+template,+**context):
++++html+=+render_to_string(f'myapp/email/{template}.html',+context+%7C+{'subject':+subject})
++++#+...+and+so+on+with+the+rest+of+my+function+for+sending+email...|1449856|entityMap^0|B|3|0|0|0|0|I|7|S|7|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

I figure what we want is an alternative to the existing <code>url</code> tag, so I wrote a new tag:
<pre><code>from django.template import Library
from django.urls import reverse

@register.simple_tag(takes_context = True)
def fullURL(context, name, *args, **kwargs):
 request = context['request']
 return f'{request.scheme}://{request.get_host()}{reverse(name, args = args, kwargs = kwargs)}'
</code></pre>
Then in your template you can have this...
<pre><code>{% extends &quot;myapp/email/email_base.html&quot; %}

{% load mytags %} {# Replace mytags with whatever the name of your custom tags calss is. #}

{% block content %}
&lt;p&gt;You can use &lt;a href=&quot;{% fullURL 'signup' %}&quot;&gt;this link&lt;/a&gt; to get started with your account. We look forward to seeing you soon!&lt;/p&gt;
{% endblock content %}
</code></pre>
Then when you generate this, you just need to remember to pass the <code>request</code> into the <code>context</code>, like so...
<pre><code>from django.template.loader import render_to_string

def sendEmail(subject, to, template, **context):
 html = render_to_string(f'myapp/email/{template}.html', context | {'subject': subject})
 # ... and so on with the rest of my function for sending email...
</code></pre>

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.

How can I get the domain name of my site within a Django template?

数据库

服务器

Django

如何从Django模板中获取当前站点的域名？我试着查看了标签和过滤器，但什么都没有。

问如何在Django模板中获取站点的域名？
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问如何在Django模板中获取站点的域名？
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