Question

我正在尝试在django的主页上列出所有具有相同类别的帖子。我希望标题成为类别标题，然后在标题下方显示所有与该类别相关的帖子。我通过在stackoverflow上寻求帮助在基于类的视图中做到了这一点，但我想在基于函数的视图中了解这一点，以了解这一点。

posts / models.py

from tinymce import HTMLField
from django.db import models
from django.contrib.auth import get_user_model
from slugger import AutoSlugField
from django.urls import reverse
# Create your models here.
User = get_user_model()


def upload_location(instance, filename):
    return "%s/%s" %(instance.slug, filename)


class Author(models.Model):
    user = models.OneToOneField(User, on_delete=models.CASCADE)

    def __str__(self):
        return self.user.username

class Category(models.Model):
    title = models.CharField(max_length=20)
    slug = AutoSlugField(populate_from='title')
    def __str__(self):
        return self.title


class Post(models.Model):
    title = models.CharField(max_length = 100)
    slug = AutoSlugField(populate_from='title')
    overview = models.CharField(max_length= 200)
    timestamp = models.DateTimeField(auto_now_add=True)
    content = HTMLField()
    comment_count = models.IntegerField(default=0)
    view_count = models.IntegerField(default=0)
    author = models.ForeignKey(Author, on_delete=models.CASCADE)
    thumbnail = models.ImageField(
        upload_to=upload_location,
        null=True, 
        blank=True)
    category = models.ManyToManyField(Category)
    featured = models.BooleanField()
    previous_post = models.ForeignKey('self', related_name= 'previous', on_delete=models.SET_NULL, blank=True, null=True)
    next_post = models.ForeignKey('self', related_name= 'next', on_delete=models.SET_NULL, blank=True, null=True)

    def __str__(self):
        return self.title

posts / views.py我尝试执行此操作，但这给我这样的错误

TypeError位于/   index（）缺少1个必需的位置参数：“ category_slug”

from django.shortcuts import render
from .models import Post, Author, Category
# Create your views here.
def index(request):
    featured = Post.objects.filter(featured = True)   #put this on carousel
    latest_post = Post.objects.order_by('-timestamp')[:6]
    category = Category.objects.filter(slug= category_slug)
    post_by_category = Post.objects.filter(category=category)
    context = {
        'object_list': featured,
        'latest_post': latest_post,
        'post_by_category': post_by_category,
    }
    return render(request, 'index.html', context)

urls.py

from django.contrib import admin
from django.urls import path, include
from django.conf.urls.static import static
from django.conf import settings
from posts.views import index
urlpatterns = [
    path('admin/', admin.site.urls),
    path('', index),
    path('tinymce/', include('tinymce.urls'))
]

if settings.DEBUG:
    urlpatterns += static(settings.STATIC_URL, document_root=settings.STATIC_ROOT)
    urlpatterns += static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

Answer 1

在使用基于函数的视图（如def index(request, category_slug):）时，您需要声明一个query_param，并且必须在URL中捕获它。试试这个：

path('<str:category_slug>', index)

作为您的索引视图

Answer 2

因此，这是您的网址，您可以从其中获取类别信息以过滤帖子

path('<str:category_slug>', index)

您的请求函数必须将url参数作为参数，以便您可以从数据库中查询类别，在这种情况下，该类别为category_slug

def index(request, category_slug):
    category = Category.objects.get(slug=category_slug)
    post_by_category = Post.objects.filter(category=category)

请记住，filter（）返回实例列表，因此请改用get（）。

如何在基于Django功能的视图中查询获取按相同类别过滤的帖子

2 个答案: