Abb*_*Med 3 html python django url slug
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，一个有一个 slug，**第一个 slug 是第一个，第二个是第二个问题是当我尝试时到达http://127.0.0.1:8000/posts/first或http://127.0.0.1:8000/posts/second它给我一个 404 错误，它告诉我**
使用 custom.urls 中定义的 URLconf，Django 按以下顺序尝试了这些 URL 模式：
Run Code Online (Sandbox Code Playgroud)admin/ posts/ [name='posts_list'] <slug>当前路径，posts/first，与这些都不匹配。
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>
Wil*_*sem 10
Aslug从不包含斜线。看起来您的网址以posts/. 所以你可以改变你urls.py的：
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'),
]添加路径转换器的类型可能会更好，因此.<slug:slug>
您可能希望使用django-autoslug[GitHub]自动构建基于特定字段的 slug。
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