【问题标题】:How to get the previous and next post in my Django blog?如何在我的 Django 博客中获取上一篇和下一篇文章?
【发布时间】:2015-06-26 10:07:40
【问题描述】:

我知道 Django 有 get_previous_by_FOO 和 get_next_by_FOO 方法,但我不知道如何在我的模板和视图中使用它们。所以,请帮助我使这两种方法适用于我的项目。

我正在运行 Django1.7Python2.7,我的应用名称是 blog 这是我对应的文件:

blog/views.py

def view_post(request, slug):
    post = get_object_or_404(Post, slug=slug)
    return render_to_response('blog/blog_post.html',
            {
                'post':post,
            },
            context_instance=RequestContext(request))



class PublishedPostMixin(object):
    def get_queryset(self):
        return self.model.objects.live()


class PostDetailView(PublishedPostMixin,DetailView):
    model = Post

blog/models.py

class Post(models.Model):
    created_at = models.DateTimeField(auto_now_add=True, editable=False)
    updated_at = models.DateTimeField(auto_now=True, editable=False)
    title = models.CharField(max_length=255)
    slug = models.SlugField(max_length=255,unique=True) 
    content = MarkdownField()
    published = models.BooleanField(default=True)
    author = models.ForeignKey(User, related_name="posts")
    tags = models.ManyToManyField(Tag)

    class Meta:
        ordering = ["-created_at", "title"]

    def __unicode__(self):
        return self.title

    def save(self, *args, ** kwargs):
        if not self.slug:
            self.slug = slugify(self.title) #title become the slug
        super(Post, self).save(*args,**kwargs)

    @models.permalink
    def get_absolute_url(self):
        return ("blog:detail",(),{'slug':self.slug
                            })

blog/urls.py

urlpatterns = patterns('',
    url(r"^(?P<slug>[\w-]+)/$",view_post, name="detail"),
)

模板

<p>
   {{ post.content|markdown }}
</p>
<ul class="pager">
   <li class="previous">
      <a href="{{post.get_previous_by_created_at}}">&larr; Previous Posts</a>
   </li>
   <li class="next">
      <a href="{{post.get_next_by_created_at}}">Next Posts &rarr;</a>
   </li>
</ul>

【问题讨论】:

标签: html django python-2.7 templates


【解决方案1】:

最后我通过在模板中引入 url 标签解决了这个问题。这是我修改后的模板。

{% load url from future %}    
<ul class="pager">
        {% if post.get_next_by_created_at %}
            <li class="previous">
                 <a href="{% url 'blog:detail' post.get_next_by_created_at.slug %}">&larr; Previous Post</a>
            </li>
        {% endif %}
        {% if post.get_previous_by_created_at %}
            <li class="next">
                 <a href="{% url 'blog:detail' post.get_previous_by_created_at.slug %}">Next Post &rarr;</a>
            </li>
        {% endif %}
    </ul>

【讨论】:

  • 对于 Django 1.11 {% load url from future %} 不再需要。
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