我有一些东西在设置。py,我想能够从模板访问,但我不知道如何做到这一点。我已经试过了
{{CONSTANT_NAME}}
但这似乎并不奏效。这可能吗?
我有一些东西在设置。py,我想能够从模板访问,但我不知道如何做到这一点。我已经试过了
{{CONSTANT_NAME}}
但这似乎并不奏效。这可能吗?
当前回答
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> ...
其他回答
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,
}
另一种方法是创建一个自定义模板标签,它可以让您从设置中获取值。
@register.tag
def value_from_settings(parser, token):
try:
# split_contents() knows not to split quoted strings.
tag_name, var = token.split_contents()
except ValueError:
raise template.TemplateSyntaxError, "%r tag requires a single argument" % token.contents.split()[0]
return ValueFromSettings(var)
class ValueFromSettings(template.Node):
def __init__(self, var):
self.arg = template.Variable(var)
def render(self, context):
return settings.__getattr__(str(self.arg))
然后你可以使用:
{% value_from_settings "FQDN" %}
将它打印在任何页面上,而无需跳过上下文处理器的约束。
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 }}
如果有人像我一样发现了这个问题,那么我将发布我的解决方案,它适用于Django 2.0:
这个标记将一些settings.py变量值赋给模板的变量:
用法:{% get_settings_value template_var "SETTINGS_VAR" %}
应用程序/ templatetags / my_custom_tags.py:
from django import template
from django.conf import settings
register = template.Library()
class AssignNode(template.Node):
def __init__(self, name, value):
self.name = name
self.value = value
def render(self, context):
context[self.name] = getattr(settings, self.value.resolve(context, True), "")
return ''
@register.tag('get_settings_value')
def do_assign(parser, token):
bits = token.split_contents()
if len(bits) != 3:
raise template.TemplateSyntaxError("'%s' tag takes two arguments" % bits[0])
value = parser.compile_filter(bits[2])
return AssignNode(bits[1], value)
你的模板:
{% load my_custom_tags %}
# Set local template variable:
{% get_settings_value settings_debug "DEBUG" %}
# Output settings_debug variable:
{{ settings_debug }}
# Use variable in if statement:
{% if settings_debug %}
... do something ...
{% else %}
... do other stuff ...
{% endif %}
查看Django如何创建自定义模板标签的文档:https://docs.djangoproject.com/en/2.0/howto/custom-template-tags/
我发现最简单的方法是一个自定义模板标签:
from django import template
from django.conf import settings
register = template.Library()
# settings value
@register.simple_tag
def settings_value(name):
return getattr(settings, name, "")
用法:
{% settings_value "LANGUAGE_CODE" %}