Django：匹配URL模式以在404中显示数据结果

Question

我正在尝试使用URL模式匹配在模板中动态显示数据。目标是根据组织类型显示数据，我有3个组织类型：制造商、供应商和分销商。你知道吗

因此，如果URL匹配/profiles/man_dash/**manufacturers**/，则显示所有制造商的数据。下面是我的代码：

查看

def man_org_list(request, member_type=None):
    member_type_map = {
        'manufacturers': 'Manufacturer',
        'suppliers': 'Supplier',
        'distributors': 'Distributor'
    }

    member_type = member_type_map.get(member_type, None)
    if member_type is None:
        raise Http404
    queryset = Organization.objects.filter(member__member_flag=1, member__member_type=member_type).order_by('id')
    return render(request, 'profiles/man_dash.html', {'object_list': queryset})

核心网址.py

urlpatterns = [
    path('admin/', admin.site.urls),
    path('profiles/', include('profiles.urls'))
]

网址.py

urlpatterns = [
    url(r'^$', views.org_list, name='org_list'),
    url(r'^(?P<id>\d+)/$', views.org_details, name='org_details'),
    url(r'^man_dash/<str:member_type>/', views.man_org_list, name='man_org_list')
]

我要显示的代码块：

{% for org in object_list %}
      <tr>
        <th scope="row">{{ org.id }}</th>
        <td>{{ org.org_name }}</td>
        <td>{{ org.org_type }}</td>
        {% for member in org.member.all %}
        <td>{{ member.member_flag }}</td>
        {% endfor %}
        {% for c_score in org.c_score.all %}
        <td>{{ c_score.completeness_score }}%</td>
        {% endfor %}
        <td><a href="{% url 'org_details' org.id %}" target="_blank">View</a></td>
      </tr>
  {% endfor %}

出于某种原因，我不断得到错误：

The current path, profiles/man_dash/manufacturers/, didn't match any of these.

Answer 1

url(r'^man_dash/<str:member_type>/', views.man_org_list, name='man_org_list')

您在这里混淆了新旧URL语法。<str:member_type>属于path()，而r'^...'是属于url()的正则表达式。你知道吗

尝试将其更改为：

path('man_dash/<str:member_type>/', views.man_org_list, name='man_org_list')