python - ListView 缺少 QuerySet。定义 ListView.model、ListView.queryset 或覆盖 ListView.get_queryset()

Question

我真的不明白是什么导致了错误我检查了文档，这里有一个非常相似的例子是我的 views.py，我使用的应用程序下的 urls.py 包含，以及模板

View .py

class SchoolListView(ListView):
    context_object_name = 'schools'
    model = models.School

网址.py

from django.urls import path
from . import views

#My name space
app_name = 'basicapp'
urlpatterns = [
    path('', views.ListView.as_view(), name='list'),
    path('details', views.DetailView.as_view(), name='details')
]

和我的模板

{% extends 'basicapp/basicapp_base.html'%}
{% block body_block %}
  <div class="jumbotron">
    <h1>Welcome to list of all schools</h1>
    <ol>
      {% for school in schools %}
      <h2><li><a href="{{school.id}}">{{school.name}}</a></li></h2>
      {% endfor %}
    </ol>

{% endblock %}

我得到了这个我不太明白的错误

Exception Type: ImproperlyConfigured
Exception Value:    
ListView is missing a QuerySet. Define ListView.model, ListView.queryset, or override ListView.get_queryset().


Traceback Switch to copy-and-paste view
C:\ProgramData\Miniconda3\lib\site-packages\django\core\handlers\exception.py in inner
            response = get_response(request) ...
▶ Local vars
C:\ProgramData\Miniconda3\lib\site-packages\django\core\handlers\base.py in _get_response
                response = self.process_exception_by_middleware(e, request) ...
▶ Local vars
C:\ProgramData\Miniconda3\lib\site-packages\django\core\handlers\base.py in _get_response
                response = wrapped_callback(request, *callback_args, **callback_kwargs) ...
▶ Local vars
C:\ProgramData\Miniconda3\lib\site-packages\django\views\generic\base.py in view
            return self.dispatch(request, *args, **kwargs) ...
▶ Local vars
C:\ProgramData\Miniconda3\lib\site-packages\django\views\generic\base.py in dispatch
        return handler(request, *args, **kwargs) ...
▶ Local vars
C:\ProgramData\Miniconda3\lib\site-packages\django\views\generic\list.py in get
        self.object_list = self.get_queryset() ...
▶ Local vars
C:\ProgramData\Miniconda3\lib\site-packages\django\views\generic\list.py in get_queryset
                    'cls': self.__class__.__name__ ...
▶ Local vars

Answer 1

您的 urls.py 中有错误，您没有引用 SchoolListView，而是引用通用 ListView 本身。您可以通过以下方式解决此问题:

# app/urls.py

from django.urls import path
from . import views

#My name space
app_name = 'basicapp'

urlpatterns = [
    # SchoolListView instead of ListView
    path('', views.<b>SchoolListListView</b>.as_view(), name='list'),
    # <i>probably</i> SchoolDetailView instead of DetailView, and with a pk in the url
    path('details', views.DetailView.as_view(), name='details')
]

由于您在 views.py 中导入了 ListView，解释器在使用 views.ListView 时不会出错，您只需“重新- 在您的 views.py 中导出“ListView”。

可能您还定义了 SchoolDetailView 而不是 DetailView，并且 URL 可能应该包含您要显示其详细信息的学校的主键，但您确实没有提供足够的代码来解决该问题。