Question

我已经按照我给出的建议，但仍然没有在网址中显示slu ..当我尝试运行它时，我收到以下错误：

Page not found (404)
Request Method: GET
Request URL:    http://localhost:8080/blog/1/
Raised by:  blog.views.detail

我被告知是否更改了我的观点中的值

  detail(request, blog_id):
    page = "blog/detail.html"
    title = "Detail"
    context = {
      "post": get_object_or_404(Post, pk=blog_id),
   }

到这个

   detail(request, slug):
       page = "blog/detail.html"
       title = "Detail"
       context = {
        "post": get_object_or_404(Post, slug=slug),
      }

并将我的网址更改为

 url(r'^(?P<slug>[\w-]+)/$', views.detail, name='detail'),

它会起作用，但事实并非如此。下面是我的代码，现在在下面。

我在models.py中的帖子模型看起来像这样

  from django.db import models

  class Post(models.Model):
   title = models.CharField(max_length=100)
   body = models.TextField()
   slug = models.SlugField()
   img = models.CharField(max_length=100)
   created = models.DateTimeField(auto_now_add=True, auto_now=False)
   updated = models.DateTimeField(auto_now_add=False, auto_now=True)
   author = models.ForeignKey(Author)
   categories = models.ManyToManyField(Category)
   tags = models.ManyToManyField(Tag)

   def __str__(self):
       return self.title

   class Meta:
       ordering = ["-created"]

我的views.py看起来像这样

from django.shortcuts import render, get_object_or_404
from .models import Post

def detail(request, slug):
   page = "blog/detail.html"
   title = "Detail"
   context = {
      "post": get_object_or_404(Post, slug=slug),
      "title": title,
   }
   return render(request, page, context)

我的博客/ urls.py看起来像这样

from django.conf.urls import url
app_name = 'blog'

from . import views
urlpatterns = [
    # ex: /blog/
    url(r'^$', views.index, name='index'),

    # ex: /blog/5/
    url(r'^(?P<slug>[\w-]+)/$', views.detail, name='detail'),

    # ex: /blog/contact
    url(r'^contact$', views.contact, name='contact'),
]

我的网站/ urls.py看起来像这样

from django.conf.urls import include, url
from django.contrib import admin

urlpatterns = [
    url(r'^blog/', include('blog.urls')),
    url(r'^admin/', admin.site.urls),
    url(r'^$', 'blog.views.land'),
]

我的admin.py看起来像这样

from django.contrib import admin

from .models import Post, Author, Category, Tag


class PostAdmin(admin.ModelAdmin):
    prepopulated_fields = {"slug": ("title",)}
    search_fields = ['title', 'body']
    list_display = ['title', 'body', 'created']
    list_filter = ['created']

admin.site.register(Post, PostAdmin)

邀请并欢迎任何和所有帮助指导建议或教程。感谢

Answer 1

正如我在帖子中看到的那样：您尝试了网址http://localhost:8080/blog/1/，但这是旧网址，您需要新的网址，例如http://localhost:8080/blog/slug/ ...

您还应将get_object_or_404(Post, slug)更改为get_object_or_404(Post, slug=slug)。

有关详细信息，请参阅https://docs.djangoproject.com/en/1.9/_modules/django/shortcuts/#get_object_or_404，获取预期关键字参数，而不是位置：https://docs.djangoproject.com/en/1.9/ref/models/querysets/#django.db.models.query.QuerySet.get

Answer 2

检查你是否真的有一个slug 1的帖子。在我看来，您使用/blog/1来获取pk的帖子，所以现在它不会起作用，因为帖子1根本不存在

Answer 3

正则表达式是这个

(?P<slug>[\w-]+)

什么时候应该是这个

(?P<slug>[\w\-]+)

感谢所有试图提供帮助的人。我很高兴我没有放弃。这让我开启了新的一年！

Django slug抛出一个错误，网址不匹配

3 个答案: