如何不在 Python、Django 中生成两次相同的动态图像？

Question

我有一个返回生成图像的视图。我使用了两次视图：在页面上显示带有其他内容的图像，并在单击图像时在单独的窗口中显示图像。但是由于图像的生成需要很长时间，我想以某种方式逃避重复自己的需要。

views.py

# named 'dynamic-image' in urls.py
def dynamic_image(request, graph_name):
    # generate the image (always the same for our case)
    return HttpResponse(image, content_type="image/svg+xml")

def index(request):
    template_name = 'graphs/index.html'
    ...
    return render(request,template_name,{})

index.html

<a href = "{% url 'dynamic-image' simplified_cell_states_graph %}">
    <img src="{% url 'dynamic-image' simplified_cell_states_graph %}" alt="img3">
</a>

我希望我可以重用为索引模板生成的图像，通过在单独的窗口中显示它然后忘记图像。

upd按照建议添加了cache_page

@cache_page(60 * 15)
def dynamic_image(request, graph_name):
    # generate the image (always the same for our case)
    return HttpResponse(image, content_type="image/svg+xml")

upd 如何取消缓存页面？ cache.delete(key) 没有为我清除缓存。这是演示它的测试：

from django.utils.cache import get_cache_key
from django.core.cache import cache

def test_cache_invalidation_with_cache(self):
    self.factory = RequestFactory()        
    url = reverse('dynamic-image', args=('simplified_cell_states_graph',))
    request = self.factory.get(url)
    response = self.client.get(url)
    cache_key = get_cache_key(request)
    self.assertFalse(cache_key == None) #key found

    cache.delete(cache_key) # but no deletion

    cache_key = get_cache_key(request)
    self.assertEquals(cache_key, None) # fails here

Answer 1

使用cached view：

from django.views.decorators.cache import cache_page

@cache_page(60 * 15)
def dynamic_image(request, graph_name):
    ...