Django: how to generate a unique URL for a news item?

Question

Django

I am learning Django. In general, the task is typical, but so far sensible options do not come to mind. There is a website, it has news, you need to somehow generate a unique URL for each of them. Presumably, this can be done based on the date + some other unique identifier, but I can’t decide how it should work: to form a URL somehow using the template system, or to store a URL for each news in the database and substitute it in the template . Also, I still have not decided how to provide this very “uniqueness”.

Somebody will prompt the normal decision - implementation of the task? Well, or again, mb somewhere you can take the source of some kind of blog on Django to dig?

Answer 1

I did this (we refer to the news 20120221-1 - date-id):
url.py -
url(r'^news/(?P(\d){8})?(\-)?(?P(\d)+)?$', news_views.show_news, name="news-url"),
models.py:
class News(models.Model):
...........
def get_absolute_url(self):
return "/news/%s" % (self.news_date.strftime('%Y%m%d') + '-' + str(self.id))
template.html:
{{ news.title }}

Answer 2

If you like CNC and the idea of ​​breadcrumbs, then /year/month/day/id-slug scheme is preferable? for DetailView. The user can always view news for a certain period, ListView, simply by modifying the url in the browser line, in a completely logical way.

Answer 3

As far as I understand, you are interested in the question that a header-based slug may not be unique and needs to be suffixed (if it already exists in the database). The first thing that came to hand is bitbucket.org/neithere/django-autoslug
In general, www.google.com/search?hl=ru&q=django+autoslug

Answer 4

I have something like this

class Post(models.Model):
    title = models.CharField(max_length=64, verbose_name='Заголовок поста')
    uid = models.CharField(max_length=64, unique=True, blank=True, verbose_name='Url')
...........
    def get_absolute_url(self):
        return '/blog/post/%s.html' % self.uid

    def save(self, *args, **kwargs):
        if not self.uid:
            lastid = Post.objects.latest('id')
            self.uid = str(lastid.id + 1 ) +'-'+ createUid(self.title)
        super(Post, self).save(*args, **kwargs)

in functions.py file

import trans

def createUid(uid):
  return uid.encode('trans').replace(' ', '_').lower();

The url is like this
/blog/post/149-eroticheskie_gifki.html

Answer 5

Model.SlugField, django.template.defaultfilters.slugify

Answer 6

I just generated uuid