Python 错误："；不能将“无”用作查询值；在Django中尝试使用ListView分页时

我想在搜索结果中实现分页。搜索后，我看到了良好的结果（例如）

但是，当我单击“下一步”时，我有一个错误：

ValueError位于/search/，请求URL:

不能将“无”用作查询值

我认为URL（search_results.html）中存在这个问题。 我怎样才能修好它？ 我怎样才能改变：

<a href="/search?city={{ page_obj.next_page_number }}">next</a>

视图.py

from django.db import models

class City(models.Model):
    name = models.CharField(max_length=255)
    state = models.CharField(max_length=255)

    class Meta:
      verbose_name_plural = "cities"

    def __str__(self):
        return self.name

class HomePageView(ListView):
    model = City
    template_name = 'cities/home.html'
    paginate_by = 3
    page_kwarg = 'city'

def city_detail(request, pk):
    city = get_object_or_404(City, pk=pk)
    return render(request, 'cities/city_detail.html', {'city': city})


class SearchResultsView(ListView):
    model = City
    template_name = 'cities/search_results.html'
    paginate_by = 3
    page_kwarg = 'city'

    def get_queryset(self): # new
        query = self.request.GET.get('q')
        object_list = City.objects.filter(
            Q(name__icontains=query) | Q(state__icontains=query)
        )
        return object_list

urlpatterns = [
    path('search/', SearchResultsView.as_view(), name='search_results'),
    path('', HomePageView.as_view(), name='home'),
    path('city/<int:pk>/', views.city_detail, name='city_detail'),
]

url.py

from django.db import models

class City(models.Model):
    name = models.CharField(max_length=255)
    state = models.CharField(max_length=255)

    class Meta:
      verbose_name_plural = "cities"

    def __str__(self):
        return self.name

class HomePageView(ListView):
    model = City
    template_name = 'cities/home.html'
    paginate_by = 3
    page_kwarg = 'city'

def city_detail(request, pk):
    city = get_object_or_404(City, pk=pk)
    return render(request, 'cities/city_detail.html', {'city': city})


class SearchResultsView(ListView):
    model = City
    template_name = 'cities/search_results.html'
    paginate_by = 3
    page_kwarg = 'city'

    def get_queryset(self): # new
        query = self.request.GET.get('q')
        object_list = City.objects.filter(
            Q(name__icontains=query) | Q(state__icontains=query)
        )
        return object_list

urlpatterns = [
    path('search/', SearchResultsView.as_view(), name='search_results'),
    path('', HomePageView.as_view(), name='home'),
    path('city/<int:pk>/', views.city_detail, name='city_detail'),
]

urlpatterns=[
路径（'search/'，SearchResultsView.as_view（），name='search_results'），
路径（“”，HomePageView.as_view（），name='home'），
路径（'city/'，views.city\u detail，name='city\u detail'），
]

搜索结果.html

<ul>
  {% for city in object_list %}
    <li>
      {{ city.name }}, {{ city.state }}
    </li>
  {% endfor %}
</ul>

<div class="pagination">
    <span class="page-links">
        {% if page_obj.has_previous %}
            <a href="/search?city={{ page_obj.previous_page_number }}">previous</a>
        {% endif %}
            <span class="page-current">
                Page {{ page_obj.number }} of {{ page_obj.paginator.num_pages }}.
            </span>
        {% if page_obj.has_next %}
            <a href="/search?city={{ page_obj.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>

<form action="{% url 'search_results' %}" method="get">
  <input name="q" type="text" placeholder="Search...">
</form>

<ul>
  {% for city in object_list %}
    <li>
      <h1><a href="{% url 'city_detail' pk=city.pk %}">{{ city.name }}</a></h1>
    </li>
  {% endfor %}
</ul>

{对象列表%中城市的百分比}

{{city.name}，{{city.state}

{%endfor%}

{%如果页面_obj.has_previous%}
{%endif%}
第{{Page_obj.paginator.num_pages}页中的第{{Page_obj.number}页。
{%如果页面_obj.has_next%}
{%endif%}

home.html

<ul>
  {% for city in object_list %}
    <li>
      {{ city.name }}, {{ city.state }}
    </li>
  {% endfor %}
</ul>

<div class="pagination">
    <span class="page-links">
        {% if page_obj.has_previous %}
            <a href="/search?city={{ page_obj.previous_page_number }}">previous</a>
        {% endif %}
            <span class="page-current">
                Page {{ page_obj.number }} of {{ page_obj.paginator.num_pages }}.
            </span>
        {% if page_obj.has_next %}
            <a href="/search?city={{ page_obj.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>

<form action="{% url 'search_results' %}" method="get">
  <input name="q" type="text" placeholder="Search...">
</form>

<ul>
  {% for city in object_list %}
    <li>
      <h1><a href="{% url 'city_detail' pk=city.pk %}">{{ city.name }}</a></h1>
    </li>
  {% endfor %}
</ul>

{对象列表%中城市的百分比}


{%endfor%}

您会收到一个错误，提示ValueError，query为None，因为您没有在href中为您的

下一个
和
上一个
锚定标记传递查询
q

通过定义
next
和
previous
锚定标记修改search_results.html，如下所示：

<a href="/search?city={{ page_obj.next_page_number }}&q={{ query }}">next</a>

这样，
query
将被传递到模板中，以便
next
链接看起来像
http://localhost:8000/search/?city=2&q=a
修改搜索结果.html

<ul>
  {% for city in object_list %}
    <li>
      {{ city.name }}, {{ city.state }}
    </li>
  {% endfor %}
</ul>

<div class="pagination">
    <span class="page-links">
        {% if page_obj.has_previous %}
            <a href="/search?city={{ page_obj.previous_page_number }}">previous</a>
        {% endif %}
            <span class="page-current">
                Page {{ page_obj.number }} of {{ page_obj.paginator.num_pages }}.
            </span>
        {% if page_obj.has_next %}
            <a href="/search?city={{ page_obj.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>

<form action="{% url 'search_results' %}" method="get">
  <input name="q" type="text" placeholder="Search...">
</form>

<ul>
  {% for city in object_list %}
    <li>
      <h1><a href="{% url 'city_detail' pk=city.pk %}">{{ city.name }}</a></h1>
    </li>
  {% endfor %}
</ul>

更改模板中的下一页和上一页链接，如下所示



修改URL.py

from django.db import models

class City(models.Model):
    name = models.CharField(max_length=255)
    state = models.CharField(max_length=255)

    class Meta:
      verbose_name_plural = "cities"

    def __str__(self):
        return self.name

class HomePageView(ListView):
    model = City
    template_name = 'cities/home.html'
    paginate_by = 3
    page_kwarg = 'city'

def city_detail(request, pk):
    city = get_object_or_404(City, pk=pk)
    return render(request, 'cities/city_detail.html', {'city': city})


class SearchResultsView(ListView):
    model = City
    template_name = 'cities/search_results.html'
    paginate_by = 3
    page_kwarg = 'city'

    def get_queryset(self): # new
        query = self.request.GET.get('q')
        object_list = City.objects.filter(
            Q(name__icontains=query) | Q(state__icontains=query)
        )
        return object_list

urlpatterns = [
    path('search/', SearchResultsView.as_view(), name='search_results'),
    path('', HomePageView.as_view(), name='home'),
    path('city/<int:pk>/', views.city_detail, name='city_detail'),
]

从搜索结果中删除“/”，如下所示

path（'search'，SearchResultsView.as_view（），name='search_results'），

然后你的搜索URL将是这样的

http://localhost:8000/search?q=abcd&city=2
您是否可能试图在搜索视图中获取查询参数
q
，但在调用
a href
中的端点时，实际使用
city
query\u参数调用端点。当你试图获得
q
时，你一个也得不到。我不明白，你为什么要把页码作为城市的pk传递？