Question

现在我的网址是：https://mywebsite.com/newscategory_lists-5

我想将其更改为：https://mywebsite.com/categoryname

这是我的模特：

class Category(models.Model):
    name = models.CharField(max_length=40)  # 分类名

    class Meta:
        verbose_name = "分类"
        verbose_name_plural = verbose_name

    def __str__(self):
        return self.name

这是urls.py：

path('category_lists-<int:category_pk>', views.categoryNewsList, name="category_news_list"),

这是我的view.py：

def categoryNewsList(request, category_pk):
    category = get_object_or_404(Category, pk=category_pk)
    news_list = News.objects.filter(category=category)

    return render(request, "categories_list.html", {
        'news_list': news_list,
        'category': category
    })

Answer 1

尝试一下，在url.py

中

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

view.py

def categoryNewsList(request, link):
categories = {
    "graphics-design": "GD",
    "digital-marketing": "DM",
    "video-animation": "VA",
    "music-audio": "MA",
    "programming-tech": "PT"
}
try:
    foos = Category.objects.filter(name=categories[link])
    return render(request, 'yoursite/home.html', locals())
except KeyError:
    return redirect('yoursite/home.html')

，然后在您的template.html上放置链接：

    <nav class="navbar navbar-expand-sm bg-light navbar-light justify-content-center">
  <ul class="navbar-nav">
    <li class="nav-item">
      <a class="nav-link" href="/">All categories</a>
    </li>
    <li class="nav-item">
      <a class="nav-link" href="/graphics-design">Graphics & Design</a>
    </li>
    <li class="nav-item">
      <a class="nav-link" href="/digital-marketing">Digital Marketing</a>
    </li>
    <li class="nav-item">
      <a class="nav-link" href="/video-animation">Video & Animation</a>
    </li>
    <li class="nav-item">
      <a class="nav-link" href="/music-audio">Music & Audio</a>
    </li>
    <li class="nav-item">
      <a class="nav-link" href="/programming-tech">Programming & Tech</a>
    </li>
  </ul>

如何在Django中使用类别名称作为url后缀

1 个答案: