如何将复杂的 url 传递给 django 中的通用类更新视图

Question

我正在尝试为我的 django 博客项目创建更新视图，但我无法弄清楚。我有一个模型，它根据发布日期和标题创建一个 url，该标题也通过一个随机 slug 生成器给出，我无法将该 url 传递给更新视图我不断收到错误“AttributeError at /posts2020 /7/24/hello-93ej/更新/ 必须使用对象 pk 或 URLconf 中的 slug 调用通用详细视图 PostUpdateView"

这是我的代码

models.py

class Post(models.Model):
    STATUS_CHOICES = (
        ('cleared','Cleared'),('UnderReview','Being Reviewed'),('banned','Banned'),)
    title = models.CharField(max_length = 300)
    slug = models.SlugField(max_length = 300, unique_for_date='publish')
    author = models.ForeignKey(User, on_delete=models.SET_NULL, related_name='forum_posts',null=True)
    body = models.TextField()
    publish = models.DateTimeField(default=timezone.now)
    created = models.DateTimeField(auto_now_add=True)
    updated = models.DateTimeField(auto_now=True)
    status = models.CharField(max_length=12,choices=STATUS_CHOICES,default='cleared')
    objects = models.Manager()
    cleared = PublishedManager()
    class Meta:
        ordering =('-publish',)
    def __str__(self):
        return self.title
    def get_absolute_url(self):
        return reverse('posts:post_detail', args=[self.publish.year, self.publish.month, self.publish.day, self.slug])

urls.py

from . import views
from django.urls import path, include
from django.contrib.auth import views as auth_views
from .views import PostListView, PostCreateView,PostUpdateView
app_name = 'posts'
urlpatterns = [
    path('', views.PostListView.as_view(), name='post_list'),
    path('<int:year>/<int:month>/<int:day>/<slug:post>/',views.post_detail,name='post_detail'),
    path('post/new/',PostCreateView.as_view(), name='post-create'),
    path('<int:year>/<int:month>/<int:day>/<slug:post>/update/',PostUpdateView.as_view(), name='post-update'),

views.py

class PostUpdateView(LoginRequiredMixin, UpdateView):
    model = Post
    fields = ['title','body']

    def get_success_url(self):
        return reverse('posts:post-update', args=[self.publish.year, self.publish.month, self.publish.day, self.slug])
    
    def form_valid(self, form):
        form.instance.author = self.request.user
        return super().form_valid(form)

post-update.html

{% extends "Main/Base.html" %}
{% block title %} Update a post {% endblock %}
{% block content %}
    {% if request.user.is_authenticated %}
    <h1> Update a post <h1>
    <p>You can Update your post using the following form:</p>
    <form method="post">
        {{ form.as_p }}
        {% csrf_token %}
        <p><input type="submit" value="Update"></p>
    </form>
    {% endif %}
{% endblock %}

Answer 1

您可以通过覆盖get_object来过滤对象：

from django.shortcuts import get_object_or_404

class PostUpdateView(LoginRequiredMixin, UpdateView):
    model = Post
    fields = ['title','body']

    def get_object(self, *args, **kwargs):
        return get_obect_or_404(
            Post,
            publish__year=self.kwargs['year'],
            publish__month=self.kwargs['month'],
            publish__day=self.kwargs['day'],
            slug=self.kwargs['post'],
            author=self.request.user
        )

    def get_success_url(self):
        return reverse(
            'posts:post-update',
            args=[
                self.object.publish.year,
                self.object.publish.month,
                self.object.publish.day,
                self.object.slug
            ]
        )
    
    def form_valid(self, form):
        form.instance.author = self.request.user
        return super().form_valid(form)

author=self.request.user 过滤器部分将确保如果登录用户不是作者，它将引发 HTTP 404 响应，而不是让该用户编辑帖子。