Question

我正在学习django。呈现提供程序页面时出现上述错误，如果我删除了html代码，则该页面将加载而没有错误，但是使用以下代码，它将给出上述错误
预先感谢

html：

[ { "_id" : 11 }, { "_id" : 34 }, ... ]

视图：



    <ul class="all-cat-list">

                        {% for category in category_list %}

                        <li><a href="{% url 'providers: provider_list_category' category.slug %}">{{category}}
                            <span class="num-of-ads">({{category.total_providers}})</span></a></li>
                        {% endfor%}

                    </ul>
                </div>
                <ol class="breadcrumb" style="margin-bottom: 5px;">
                  <li><a href="/">Home</a></li>
                    <li class="active"><a active href="{% url 'providers :provider_list' %}> All Categories </a> </li>

                    {% if category %}

                        <li class="active">{{category}} </li>
                    {% endif%}
                </ol>
                <div class="ads-grid">
                    <div class="side-bar col-md-3">
                        <div class="search-hotel">
                        <h3 class="sear-head">Search</h3>
                        <form method="GET" action="{url 'providers : provider_list' %}">
                            <input type="text" value="Product name..."  name ="q" onfocus="this.value = '';" onblur="if (this.value == '') {this.value = 'Product name...';}" required="">
                            <input type="submit" value=" ">
                        </form>
                    </div>

网址：


    def providerlist(request,category_slug = None):
        category = None
        providerlist = Provider.objects.all()
        categorylist = Category.objects.annotate(total_providers=Count('provider'))

        if category_slug:
            category= get_object_or_404(Category, slug = category_slug)
            providerlist = Category.filter(category=category)
        search_query = request.GET.get('q')
        if search_query :
            providerlist = providerlist.filter(
                Q(name__icontains=search_query) |
                Q(description__icontains=search_query)|
                Q(category__category__name__icontains  = search_query)
        )

    template = 'provider/provider_list.html'

    context = {'provider_list': providerlist ,'category_list' : categorylist}

    return  render(request,template,context)


    def providerdetail(request,provider_slug):
        #print(provider_slug)
        providerdetail=get_object_or_404(Provider,slug=provider_slug)
        providerimage =  ProviderImages.objects.filter(provider = providerdetail)
        template ='provider/provider_detail.html'
        context = {'provider_detail': providerdetail, 'productimage' : providerimage}
        return render(request, template, context)

Answer 1

根据您在urls.py中的应用名称：

app_name = 'provider'

您必须更改所有url模板标记，并从提供程序的末尾删除s。它必须与您的app_name完全相同。如下所示：

<li><a href="{% url 'provider: provider_list_category' category.slug %}">

您还可以将应用名称更改为：

app_name = 'providers'

Answer 2

您的html文件应为<a href="{% url 'provider: provider_list_category' category.slug %}">，并确保您project_name文件夹中的urls.py具有

urlpatterns = [ path('provider', include('provider.urls', namespace='provider')),

Answer 3

您的app_name provider并将名称放在html页面providers中，您更改了app_name='providers'或在html页面中更改了

 <li class="active"><a active href="{% url 'provider:provider_list' %}"> All Categories </a> </li>

和放置内容名称空间的主urls.py，其中包含您的应用程序网址

urlpatterns = [path('provider', include('provider.urls', namespace='provider'))],

如果您更改app_name ='providers'然后更改名称空间

urlpatterns = [path('provider', include('provider.urls', namespace='providers'))],

/ provider /'providers'上的Django NoReverseMatch不是注册的名称空间

3 个答案: