Django NoReverseMatch 使用模板（再次）

Question

第二次发布关于 NoReverseMatch 的信息。我似乎无法很好地理解错误，无法在我的代码中找到错误。上次是因为我没有给我的 url 正则表达式中的字段命名，但现在这不是问题所在。我知道有一百万件事可能会出错，但由于某种原因我似乎找不到它。

错误是：

NoReverseMatch 在 /david/Physics/a/ 使用关键字参数 '{'category_name': 'Physics', 'information_name': 'Test info', 'subcategory_name': 'a'}' 反向查找。尝试了 1 种模式：['david/(?P[a-zA-Z_]+)/(?P[a-zA-Z_]+)/(?P[a-zA-Z_]+)/ $']

导致错误的模板是：

<h1>The items for {{ category }}/{{ subcategory }}</h1>
{% if items %}
    <ul>
    {% for item in items %}
        <li><a href="{% url 'information' category_name=category subcategory_name=subcategory information_name=item %}">{{ item }}</a></li>
    {% endfor %}
    </ul>
{% else %}
    <h4>No items for that subcategory</h4>
{% endif %}

views.py：

from django.shortcuts import render, get_list_or_404, get_object_or_404
from django.http import HttpResponse 

from .models import Category, Subcategory, Information

# Create your views here.

def index(request):
    return render(request, 'basic_web/index.html')

def search(request):
    return HttpResponse('Here you can search!')

def browse(request):
    categories = Category.objects.all()
    context = {'categories': categories}
    return render(request, 'basic_web/browse.html', context)

def view_category(request, category_name):
    category = get_object_or_404(Category,name__iexact=category_name)
    subcategories = get_list_or_404(Subcategory, parent=category)
    context = {'category': str(category), 'subcategories': map(str, subcategories)}
    return render(request, 'basic_web/category.html', context)

def view_subcategory(request, category_name, subcategory_name):
    category = get_object_or_404(Category,name__iexact=category_name)
    subcategory = get_object_or_404(Subcategory, name__iexact=subcategory_name, parent=category)
    items = get_list_or_404(Information, subcategory=subcategory, category=category)
    context = {'category': str(category), 'subcategory': str(subcategory), 'items': map(str,items)}
    return render(request, 'basic_web/subcategory.html', context)

def view_information(request, category_name, subcategory_name, information_name):
#    category = get_object_or_404(Category,name__iexcat=category_name)
#    subcategory = get_object_or_404(Subcategory, name__iexact=subcategory_name, parent=category)
#    information = get_object_or_404(Information, name__iexact=information_name, parent=subcategory)
    return HttpResponse('{0} {1} {2}'.format(category_name, subcategory_name, information_name))

urls.py:

from django.conf.urls import url
from . import views

urlpatterns = [
    url(r'^$', views.index, name='index'),
    url(r'^search/$', views.search, name='search'),
    url(r'^browse/$', views.browse, name='browse'),
    url(r'^(?P<category_name>[a-zA-Z_]+)/$', views.view_category, name='category'),
    url(r'^(?P<category_name>[a-zA-Z_]+)/(?P<subcategory_name>[a-zA-Z_]+)/$',
        views.view_subcategory, name='subcategory'),
    url(
    r'^(?P<category_name>[a-zA-Z_]+)/(?P<subcategory_name>[a-zA-Z_]+)/(?P<information_name>[a-zA-Z_]+)/$',
        views.view_information, name='information'),
    ]

如果您能给我一些提示，告诉我当 Django 说 NoReverseMatch 时如何识别实际错误，我将不胜感激

非常感谢！

Answer 1

这是您的 URL 定义。

r'^(?P<category_name>[a-zA-Z_]+)/(?P<subcategory_name>[a-zA-Z_]+)/(?P<information_name>[a-zA-Z_]+)/$',
    views.view_information, name='information'),

据我所知，网址中不允许有空格。但你显然在你的价值观中确实有一个空间

'{'category_name': 'Physics', 'information_name': 'Test info', 'subcategory_name': 'a'}'