在一个视图中显示多个模型，并获取每个后django的slug url

from django.db import models
from django.urls import reverse

class Category(models.Model):
    name = models.CharField(max_length=100)
    slug = models.SlugField(unique=True)

    def get_absolute_url(self):
        return reverse('category_detail', args=[self.slug])

class Post(models.Model):
    title = models.CharField(max_length=100)
    content = models.TextField()
    slug = models.SlugField(unique=True)
    category = models.ForeignKey(Category, on_delete=models.CASCADE)

    def get_absolute_url(self):
        return reverse('post_detail', args=[self.slug])

from django.views.generic import ListView

class PostCategoryView(ListView):
    model = Post
    template_name = 'post_category.html'
    context_object_name = 'posts'

    def get_queryset(self):
        category_slug = self.kwargs['category_slug']
        category = Category.objects.get(slug=category_slug)
        return Post.objects.filter(category=category)

{% for post in posts %}
    <h2><a href="{{ post.get_absolute_url }}">{{ post.title }}</a></h2>
    <p>{{ post.content }}</p>
{% endfor %}

from django.urls import path
from .views import PostCategoryView

urlpatterns = [
    path('category/<slug:category_slug>/', PostCategoryView.as_view(), name='category_detail'),
]