Django:匹配URL模式以显示404中的数据

Question

我正在尝试使用URL模式匹配来在模板中动态显示数据。目标是显示基于组织类型的数据，我有3个组织类型：制造商、供应商和分销商。

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

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urls.py

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

urls.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，我一直收到错误：

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。

尝试将其更改为：

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