python - 使用 ListView django 进行分页

Question

我想创建带有搜索和分页的应用程序。分页不适用于 ListView。

当我单击“下一步”链接时，我将从起始页面http://127.0.0.1:8001/ ---> 移动到 http://127.0.0.1:8001/?city=2但列表的元素没有改变。

然后单击“下一步”链接并没有更改网址（http://127.0.0.1:8001/?city=2 --> http://127.0.0.1:8001/?city=2）。

你能帮我找出错误吗？ 我认为 *.html 文件中的错误，但找不到

我的代码： models.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

网址.py

# cities/urls.py
from django.urls import path
from . import views
from .views import HomePageView, SearchResultsView

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'),
]

视图.py

from django.shortcuts import render
from django.views.generic import TemplateView, ListView
from .models import City
from django.db.models import Q
from django.shortcuts import render, get_object_or_404


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

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'

    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

主页.html

<!-- templates/home.html -->
<h1>HomePage</h1>

<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>


{{page_obj}}

<div class="pagination">
    <span class="page-links">
        {% if page_obj.has_previous %}
            <a href="/?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="/?city={{ page_obj.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>

标签： pythondjangopython-3.x

Answer 1

页面查询参数的标准名称是'page'您应该更改查询参数的名称，或者使用该?page=参数呈现模板。

选项 1：更改page_kwarg

您可以通过更改page_kwarg属性 [Django-doc]来更改它：

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

选项 2：更改模板

然而，简单地更改模板可能更有意义，以便它page用作参数：

<div class="pagination">
    <span class="page-links">
        {% if page_obj.has_previous %}
            <a href="/?page={{ 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="/?page={{ page_obj.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>