NoReverseMatch错误Django与slug

Question

我试图让索引视图显示一个＆＃34; Journal＆＃34; s列表，每个链接到列出＆＃34; Manuscripts＆＃34;在期刊上。我还希望journal_view位于期刊名称的slug中，同样适用于manuscript_view。

这是我得到的追溯：

Environment:
Request Method: GET
Request URL: http://127.0.0.1:8000/journals/

Django Version: 2.0.3
Python Version: 3.6.4
Installed Applications:
['social_django',
 'django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'journal.apps.JournalConfig']
Installed Middleware:
['django.middleware.security.SecurityMiddleware',
 'django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware']


Template error:
In template ... error at line 8
   Reverse for 'view_journal' not found. 'view_journal' is not a valid view function or pattern name.
   1 : {% load static %}
   2 : 
   3 : <link rel="stylesheet" type="text/css" href="{% static 'journal/style.css' %}" />
   4 : 
   5 : {% if journal_list %}
   6 :     <ul>
   7 :     {% for journal in journal_list %}
   8 :         <li><a href=" {% url journal.get_absolute_url journal.slug %} ">{{ journal.title }}</a></li>
   9 :     {% endfor %}
   10 :     </ul>
   11 : {% else %}
   12 :     <p>No journals are available.</p>
   13 : {% endif %}

# ...(Traceback)...

    Exception Type: NoReverseMatch at /journals/
    Exception Value: Reverse for 'view_journal' not found. 'view_journal' is not a valid view function or pattern name.

Journal模型如下所示：

class Journal(models.Model):
    title = models.CharField(max_length=100, unique=True)
    slug = models.SlugField(unique=True)
    # blah blah

    def __str__(self):
        return self.title

    def get_absolute_url(self, *args, **kwargs):
        return reverse("view_journal", kwargs={"slug": self.slug})

我的观点如下：

class IndexView(generic.ListView):   
    template_name = 'journal/index.html'
    context_object_name = 'journal_list'

    def get_queryset(self):
        return Journal.objects.all()


class JournalView(generic.ListView):
    template_name = 'journal/journal.html'
    context_object_name = 'latest_article_list'

    def get_queryset(self):
        return Manuscript.objects.filter(
                                    pub_date__lte=timezone.now()
                                        ).order_by('-pub_date')[:5]

我的应用网址如下所示：

app_name = 'journal'
urlpatterns = [
    path('', views.IndexView.as_view(), name='index'),
    path('<slug>/', views.JournalView.as_view(), name='view_journal'),
    path('<slug>/', views.ArticleView.as_view(), name='view_manuscript'), ]

项目网址：

from django.contrib import admin
from django.urls import path, include

urlpatterns = [
    path('admin/', admin.site.urls),
    path('journals/', include('journal.urls')),
    path('social/', include('social.apps.django_app.urls', namespace='social')),
]

Answer 1

我以这种方式工作：

我首先在我的模型中定义了一个slug字段，正如你所做的那样：

greater<int>

然后在我的views.py中我添加了这个：

slug = models.SlugField(max_length=20, unique=True)

最后用这种方式写了我的路径：

slug_url_kwarg = 'slug'