你好，我是Django的新手，我正在尝试建立一个网站。在管理页面http://127.0.0.1:8000/admin/posts/post/，我添加了两个帖子，一个是have，**第一个是第一，第二个是第二问题是，当我尝试到达http://127.0.0.1:8000/posts/first或http://127.0.0.1:8000/posts/second时，它给我一个404错误，它告诉我**
[使用custom.urls中定义的URLconf，Django尝试了这些URL模式，按此顺序：
admin/ posts/ [name='posts_list'] <slug>当前路径，帖子/第一，与任何这些都不匹配。
这是models.py
from django.db import models
from django.conf import settings
Create your models here.
User = settings.AUTH_USER_MODEL
class Author(models.Model):
user = models.ForeignKey(User, on_delete=models.CASCADE)
email = models.EmailField()
phone_num = models.IntegerField(("Phone number"))
def __str__(self):
return self.user.username
class Post(models.Model):
title = models.CharField(max_length=120)
description = models.TextField()
slug = models.SlugField()
image = models.ImageField()
author = models.OneToOneField(Author, on_delete=models.CASCADE)
def __str__(self):
return self.title
这是views.py
from django.shortcuts import render, get_object_or_404
from .models import Post
# Create your views here.
def posts_list(request):
all_posts = Post.objects.all()
return render(request,
"posts/posts_list.html",
context = {"all_posts": all_posts})
def posts_detail(request, slug):
unique_slug = get_object_or_404(Post, slug = slug)
return render(request, "posts/posts_detail.html", {"post": unique_slug})
这是urls.py
from django.contrib import admin
from django.urls import path
from posts.views import posts_list, posts_detail
urlpatterns = [
path('admin/', admin.site.urls),
path("posts/", posts_list, name = "posts_list"),
path("<slug>", posts_detail), #, name = "unique_slug"
]
这是这些模板：posts_list.html
<!DOCTYPE html>
<html>
<head>
<title>
</title>
</head>
<body>
{{ all_posts }}
</body>
</html>
post_detail.html
<!DOCTYPE html>
<html>
<head>
<title>
</title>
</head>
<body>
{{ post }}
</body>
</html>
slug绝不包含斜杠。您的网址似乎以posts/为前缀。因此，您可以使用以下方法更改urls.py：from django.contrib import admin
from django.urls import path
from posts.views import posts_list, posts_detail
urlpatterns = [
path('admin/', admin.site.urls),
path('posts/', posts_list, name='posts_list'),
path('posts/<slug:slug>/', posts_detail, name='unique_slug'),
]
最好添加路径转换器的类型，所以<slug:slug>。您可能想使用
django-autoslug[GitHub]根据特定字段自动构建一个条。