将基于 class 的 ListView 作为模板片段包含在 TemplateView 中

Question

我想要以下内容：

一个模板视图（实际上是我的登录页面），其中包含许多项目，包括一个列表视图。

现在，我有一个列表视图，当我将它映射到它自己的列表视图时可以使用 url：

app/views.py

class MymodelListView(ListView):
    model = Mymodel
    context_object_name = "mymodel_list"

app/urls.py

app_name = "myapp"
urlpatterns = [
 ...
    path("mlist/", MymodelListView.as_view(), name="mlist"),
 ...
]

应用/../mymodel_list.html

{% extends 'base.html' %}
{% block content %}
<ul>
    {% for adventure in adventure_list %}
    <li>
        <a href="{{ adventure.get_absolute_url }}">{{ adventure.title }}</a>
    </li>
    {% endfor %}
</ul>
{% endblock content %}

现在这按预期工作，并在 "mlist" link.

处显示来自 Mymodel 的所有记录

不过，我想将它嵌入到我的 index.html TemplateView 中。我的想法是使用一个片段并将其放入 index.html

{% include 'myapp/_mymodel_list.html' with mymodel_list=mymodel_list %}

（不确定 "with" 部分是否是必需的，但似乎我应该需要将列表从主模板传递到代码段。目前无论如何都不起作用，因为更基本的问题）。

然后我的主主页视图设置如下：

class HomePageView(TemplateView):
    template_name = "index.html"

    def get_context_data(self, *args, **kwargs):
        context = super(HomePageView, self).get_context_data(*args, **kwargs)
        context["adventure_list"] = MymodelListView.get_context_data()
        print(context)
        return context

但这会崩溃：

Exception Type: TypeError at /
Exception Value: super(type, obj): obj must be an instance or subtype of type

如果我将 self 传递给 MymodelListView.get_context_data(self)。

也一样

据我所知，Mymodel get context 很混乱，因为它被传递给了 HomePage context，而且有点不对劲。

我想做的是完全错误的吗？这就是为什么我在我的（通常相当可靠的）google-fu 中找不到任何有用的提示吗？如果是这样，我应该采取什么方法呢？

我想在其他页面中重新使用 Mymodel 列表视图（尽管可能会更改过滤条件），因此我尝试以 DRY 方式这样做。

Answer 1

您遇到的问题是您试图用 HomePageView 的实例调用 MymodelListView 的方法。因为 HomePageView 没有继承自 MymodelListView Python 不能这样做并且会抛出错误。你有几种方法可以解决这个问题。

myapp/views.py

# The first approach is to use Django's included mixin to get the
# queryset and include it in the context. This is a bit redundant and
# won't include any logic used in your ListView.

from django.views.generic.list import MultipleObjectMixin


class HomePageView(MultipleObjectMixin, TemplateView):
    context_object_name = 'adventure_list'
    model = Mymodel
    template_name = 'index.html'


# Your second option is to manually include the queryset. Again, this
# won't include any login from the ListView.

class HomePageView(TemplateView):
    template_name = 'index.html'

    def get_context_data(self, **kwargs):
        kwargs.setdefault('adventure_list', Mymodel.objects.all())
        return super().get_context_data(**kwargs)


# This is the option that I would suggest. Inherit from the ListView
# that you've already written and give it its own template and context
# name. I'm pretty sure this should do exactly what you want.

class HomePageView(MymodelListView):
    context_object_name = 'adventure_list'
    template_name = 'index.html'

我预见到的另一个问题是您正在尝试 include 整个 HTML 文件。我怀疑这对你有用。您最好将列表放在一个专门要包含的文件中，然后在两个地方都使用它。您唯一认为会复制的是 include 标签。

templates/index.html

...
<!-- If the context name doesn't change you don't need to assign it and
Django will render the included file correctly. -->
{% include 'myapp/include_list.html' %}
...

templates/myapp/mymodel_list.html

{% extends 'base.html' %}

{% block content %}
  {% include 'myapp/include_list.html' with adventure_list=mymodel_list %}
{% endblock %}

templates/myapp/include_list.html

<ul>
  {% for adventure in adventure_list %}
    <li><a href="{{ adventure.get_absolute_url }}">{{ adventure.title }}</a></li>
  {% endfor %}
</ul>

将基于 class 的 ListView 作为模板片段包含在 TemplateView 中

Include class based ListView as a template snippet in a TemplateView

django-templates

django-views

django-class-based-views

python-3.7

django-2.2