我是相当新的Django,我希望我能够正确地制定我的问题,请裸露在我身边。使用Django Logic来添加CSS类基于URL与slu 012
我有什么权利现在:
一个简单的个人网站,一个博客节。当我从导航栏转到不同的列表项目时,如果li处于活动状态,它将添加一个“激活”类。问题发生在我的博客部分。
我想实现:
当我添加了“激活”类的博客节和博客这样的链接将被激活。我用我的帖子网址slu g。
我有什么权利现在它不工作:
template.html
<li class="{% if request.resolver_match.url_name == "index" %}activate{% endif %}"><a href='/'>Home</a></li>
<li class="{% if request.resolver_match.url_name == "blog" %}activate{% endif %}"><a href='/blog/'>Blog</a></li>
<li class="{% if request.resolver_match.url_name == "photohraphy" %}activate{% endif %}"><a href='/photography/'>Photography</a></li>
models.py
from django.db import models
from django.core.urlresolvers import reverse
class Post(models.Model):
title = models.CharField(max_length=250)
slug = models.SlugField(max_length=250)
body = models.TextField()
date = models.DateTimeField()
updated = models.DateTimeField(auto_now=True)
def get_absolute_url(self):
return reverse('blog:post_detail', args=[self.slug])
def __str__(self):
return self.title
urls.py
from django.conf.urls import url, include
from django.views.generic import ListView, DetailView
from blog.models import Post
from . import views
urlpatterns = [
url(r'^$', views.list_of_post, name='list_of_post'),
url(r'^(?P<slug>[-\w]+)/$', views.post_detail, name='post_detail')
views.py
from django.shortcuts import render, get_object_or_404
from .models import Post
def list_of_post(request):
post = Post.objects.all()
template = 'blog/blog.html'
context = {'post': post}
return render(request, template, context)
def post_detail(request, slug):
post = get_object_or_404(Post, slug=slug)
template = 'blog/post.html'
context = {'post': post}
return render(request, template, context)
我知道,我必须添加一些神社逻辑在李类,但我不知道如何正确设置它。我想这与get_absolute_url有关。有人可以建议一种简单而正确的方法吗?
谢谢你这么多
请更正'urls.py',并添加'viewsy.py'如果你能:) – opalczynski
的urls.py似乎正常工作。我不明白我应该改正什么。我添加了views.py。谢谢! – IoanCosmin
我可以看到,urls.py实际上是models.py - 可能你只是复制了两次... – opalczynski