Django将ID转换为slug

Question

我是django的新手，想要将id / pk转换为slug ，我想让 url变得友好或易于理解

想要将网址http://127.0.0.1:8000/1/设为http://127.0.0.1:8000/hello/

Model.py

class Post(models.Model):
    title=models.CharField(max_length=200)
    description=models.TextField(max_length=10000)
    pub_date=models.DateTimeField(auto_now_add=True)
    comments=models.CharField(max_length=200, blank=True)
    slug = models.SlugField(max_length=40, unique=True)

    def __unicode__(self):
        return self.title

    def description_as_list(self):
        return self.description.split('\n')

admin.py

class PostAdmin(admin.ModelAdmin):
    list_display=['title','description']
    prepopulated_fields = {'slug': ('title',)}

    class Meta:
        model = Post

admin.site.register(Post,PostAdmin)

urls.py

urlpatterns = [

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

views.py

class PostListView(ListView):
    model = Post
    template_name = 'blog_post.html'
    queryset = Post.objects.order_by('-pub_date')
    paginate_by = 2

def detail(request, id):
    posts = Post.objects.get(id=id)
    return render(request, "blog_detail.html", {'posts': posts,})

模板

{% for threads in object_list  %}
    <p class="blog-post-title"><a href="{% url 'detail' slug=threads.id %}">{{ threads.title }}</a></p>
    <hr />
{% endfor %}

通过这样做可以获得错误 ...还有什么更改

任何帮助都可以使url可读。谢谢davance

Answer 1

要实现您的目标，您需要将slug参数传递给views.py的详细功能，如下所示：

def detail(request, slug):
    posts = Post.objects.get(slug=slug)
    return render(request, "blog_detail.html", {'posts': posts,})

此外，您的blog_post.html模板应该接受网址中的slug：

{% for threads in object_list  %}
    <p class="blog-post-title"><a href="{% url 'detail' slug=threads.slug %}">{{ threads.title }}</a></p>
    <hr />
{% endfor %}

当然，您需要为详细信息视图提供blog_detail模板。