我在Django中有一个视图,该视图使用备忘录来缓存数据,以获取依赖相对静态数据的更高度运输的视图。关键词是相对的:我需要在数据库中更改该特定URL数据的模因密钥无效。为了尽可能清楚,这是肉类的肉(人是模型,缓存是django.core.cache.cache.cache):

def person_detail(request, slug): 
    if request.is_ajax():
        cache_key = "%s_ABOUT_%s" % settings.SITE_PREFIX, slug

        # Check the cache to see if we've already got this result made.
        json_dict = cache.get(cache_key)

        # Was it a cache hit?
        if json_dict is None:
            # That's a negative Ghost Rider
            person = get_object_or_404(Person, display = True, slug = slug)

            json_dict = {
                'name' : person.name,
                'bio' : person.bio_html,
                'image' : person.image.extra_thumbnails['large'].absolute_url,
            }

            cache.set(cache_key)

        # json_dict will now exist, whether it's from the cache or not
        response = HttpResponse()
        response['Content-Type'] = 'text/javascript'
        response.write(simpljson.dumps(json_dict)) # Make sure it's all properly formatted for JS by using simplejson
        return response
    else:
        # This is where the fully templated response is generated

我想做的是在其“未形式”表单中获取该cache_key变量,但我不确定该怎么做 - 如果可以做到这一点。

以防万一已经有事要这样做,这是我想处理的事情(这是来自人模型的假设保存方法)

def save(self):    
    # If this is an update, the key will be cached, otherwise it won't, let's see if we can't find me
    try:
        old_self = Person.objects.get(pk=self.id)
        cache_key = # Voodoo magic to get that variable
        old_key = cache_key.format(settings.SITE_PREFIX, old_self.slug) # Generate the key currently cached
        cache.delete(old_key) # Hit it with both barrels of rock salt

    # Turns out this  doesn't already exist, let's make that first request even faster by making this cache right now
    except DoesNotExist:
        # I haven't gotten to this yet.

    super(Person, self).save()

我正在考虑为这种东西上一个视图课,并在其中具有功能 remove_cache 或者 generate_cache 既然我做了这种事情 很多. 。这是一个更好的主意吗?如果是这样,如果他们在班级中,我该如何调用urlconf中的视图?

有帮助吗?

解决方案

URLCONF应指向任何可召唤。没有严格的要求使其指向正常工作。您可以使用缓存方法实现基类,然后将其扩展:

class RealView(BaseViewWithCacheMethods):
    def __call__(self, request):
        if request.is_ajax():
            return self.ajax_view()
        return self.html_view()

URLCONF定义将是类似的:

from django.conf.urls.defaults import *
from views import RealView

urlpattrens = patterns('',
    (r'^$', RealView()),
)
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