如何不在Django中仅缓存我的模板的一段代码

Question

Under django v. 1.4.

The problem:

Since my template is rendered by this view which will be totally stored in cache:

@cache_page(60*60*24)
def index(request):
    foo_form = FooForm()
    context = RequestContext(request, {
        'foo_form': foo_form
    })
    # An entire page is rendered
    return render_to_response('index.html', context_instance=context)

In my template I have an if statement which checks whether user is authenticate:

...
<li>
    {%  if user.is_authenticated %}
        <a href="{% url 'home' %}" class="login">Enter</a>
    {% else %}
        <a href="" class="login" data-target="#login_modal" data-tggle="modal">Enter</a>
    {% endif %}

</li>
...

There is a modal that's activated by a button "Enter" which should be displayed when there's no user logged, otherwise the user is redirected to the system when the "Enter" button is clicked.

The question: Is there a way to ignore only that piece of code from my template to be not cached? If so, how?

Answer 1

You should use Template fragment caching: https://docs.djangoproject.com/en/1.6/topics/cache/#template-fragment-caching

{% load cache %}
{% cache 500 sidebar request.user.username %}
    .. sidebar for logged in user ..
{% endcache %}

Answer 2

Use ajax to display the user authentication template, so caching wont affect your template.

<li>

`{%  if user.is_authenticated %}`
   ` <a href="{% url 'home' %}" class="login">Enter</a>`
`{% else %}`
    `<a href="" class="login" data-target="#login_modal" data-tggle="modal">Enter</a>`

{% endif %}

</li>