我喜欢Berislav的解决方案，因为在简单的网站上，它干净有效。我不喜欢的是随意地暴露所有的设置常数。所以我最后是这样做的:

from django import template
from django.conf import settings

register = template.Library()

ALLOWABLE_VALUES = ("CONSTANT_NAME_1", "CONSTANT_NAME_2",)

# settings value
@register.simple_tag
def settings_value(name):
    if name in ALLOWABLE_VALUES:
        return getattr(settings, name, '')
    return ''

用法:

{% settings_value "CONSTANT_NAME_1" %}

This protects any constants that you have not named from use in the template, and if you wanted to get really fancy, you could set a tuple in the settings, and create more than one template tag for different pages, apps or areas, and simply combine a local tuple with the settings tuple as needed, then do the list comprehension to see if the value is acceptable. I agree, on a complex site, this is a bit simplistic, but there are values that would be nice to have universally in templates, and this seems to work nicely. Thanks to Berislav for the original idea!

上面来自bchhun的例子很好，只是你需要从settings.py显式地构建上下文字典。下面是一个未经测试的示例，说明如何从settings.py的所有大写属性(re: "^[A-Z0-9_]+$")自动构建上下文字典。

在settings.py的末尾:

_context = {} 
local_context = locals()
for (k,v) in local_context.items():
    if re.search('^[A-Z0-9_]+$',k):
        _context[k] = str(v)

def settings_context(context):
    return _context

TEMPLATE_CONTEXT_PROCESSORS = (
...
'myproject.settings.settings_context',
...
)

将这段代码添加到名为context_processors.py的文件中:

from django.conf import settings as django_settings


def settings(request):
    return {
        'settings': django_settings,
    }

然后，在你的设置文件中，包括一个路径，如'speed .core.base.context_processors。在TEMPLATES中的context_processors设置中的settings'(包含你的应用程序名称和路径)。

(例如，你可以看到settings/base.py和context_processors.py)。

然后可以在任何模板代码中使用特定的设置。例如:

{% if settings.SITE_ID == settings.SPEEDY_MATCH_SITE_ID %}

更新:上面的代码向模板公开了所有设置，包括敏感信息，如SECRET_KEY。黑客可能滥用此特性在模板中显示此类信息。如果你只想将特定的设置暴露给模板，请使用下面的代码:

def settings(request):
    settings_in_templates = {}
    for attr in ["SITE_ID", ...]: # Write here the settings you want to expose to the templates.
        if (hasattr(django_settings, attr)):
            settings_in_templates[attr] = getattr(django_settings, attr)
    return {
        'settings': settings_in_templates,
    }

如果使用基于类的视图:

#
# in settings.py
#
YOUR_CUSTOM_SETTING = 'some value'

#
# in views.py
#
from django.conf import settings #for getting settings vars

class YourView(DetailView): #assuming DetailView; whatever though

    # ...

    def get_context_data(self, **kwargs):

        context = super(YourView, self).get_context_data(**kwargs)
        context['YOUR_CUSTOM_SETTING'] = settings.YOUR_CUSTOM_SETTING

        return context

#
# in your_template.html, reference the setting like any other context variable
#
{{ YOUR_CUSTOM_SETTING }}

在Django 2.0+中添加了一个完整的创建自定义模板标签的答案

在你的app文件夹中，创建一个名为templatetags的文件夹。在其中，创建__init__.py和custom_tags.py:

在custom_tags.py中创建一个自定义标记函数，用于访问settings常量中的任意键:

from django import template
from django.conf import settings

register = template.Library()

@register.simple_tag
def get_setting(name):
    return getattr(settings, name, "")

要理解这段代码，我建议阅读Django文档中关于简单标记的部分。

然后，你需要在任何模板中加载这个文件，让Django知道这个(以及任何其他)自定义标记。就像你需要加载内置的静态标签一样:

{% load custom_tags %}

加载后，它可以像任何其他标签一样使用，只需提供您需要返回的特定设置。如果你有一个BUILD_VERSION变量在你的设置:

{% get_setting "BUILD_VERSION" %}

此解决方案不适用于数组，但如果需要，则可能需要在模板中放入太多逻辑。

注意:一个更清晰、更安全的解决方案可能是创建一个自定义上下文处理器，在其中添加所有模板可用的上下文所需的设置。这样可以降低在模板中错误地输出敏感设置的风险。

如果你希望每个请求和模板都有一个值，那么使用上下文处理器更合适。

方法如下:

Make a context_processors.py file in your app directory. Let's say I want to have the ADMIN_PREFIX_VALUE value in every context: from django.conf import settings # import the settings file def admin_media(request): # return the value you want as a dictionnary. you may add multiple values in there. return {'ADMIN_MEDIA_URL': settings.ADMIN_MEDIA_PREFIX} add your context processor to your settings.py file: TEMPLATES = [{ # whatever comes before 'OPTIONS': { 'context_processors': [ # whatever comes before "your_app.context_processors.admin_media", ], } }] Use RequestContext in your view to add your context processors in your template. The render shortcut does this automatically: from django.shortcuts import render def my_view(request): return render(request, "index.html") and finally, in your template: ... <a href="{{ ADMIN_MEDIA_URL }}">path to admin media</a> ...