How to Create a Slug in Django

Django: how to create slugs in django?

Creating posts with the same slug makes not much sense, since a slug is used to determine what Article is used. If two items have learning-to-code, then you can not determine which of the two Articles is the right one.

If your Article for example has a DateField, you can use this such that is works with both the date and the slug to determine the Article. The easist way to achieve that is likely with the django-autoslug package ^[PyPi]. This will work with a UniqueConstraint ^[Django-doc], such that the combination of the date and the slug is unique, so:

from autoslug import AutoSlugField
from django.conf import settings

class Article(models.Model):
    title = models.CharField(max_length=200)
    slug = AutoSlugField(populate_from='title', unique_with=['publish_date'])
    description = models.TextField(blank=True, null=True)
    user = models.ForeignKey(
        settings.AUTH_USER_MODEL,
        on_delete=models.SET_NULL,
        null=True
    )
    publish_date = models.DateField(auto_now_add=True)

The urls.py will need to add specifications for the date, so:

path('<int:year>/<int:month>/<int:date>/<slug:slug>', some_view, name='article_detail')

then in a DetailView, or other views, we can filter on the QuerySet with the date:

from datetime import date
from django.views.generic import DetailView

class ArticleDetailView(DetailView):
    model = Article
    
    def get_queryset(self, *args, **kwargs):
        return super().get_queryset(*args, **kwargs).filter(
            publish_date=date(self.kwargs['year'], self.kwargs['month'], self.kwargs['day'])
        )

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Note: It is normally better to make use of the settings.AUTH_USER_MODEL ^[Django-doc] to refer to the user model, than to use the User model ^[Django-doc] directly. For more information you can see the referencing the User model section of the documentation.

How to auto generate slug from my Album model in django 2.0.4

Django autogenerates the slug from the string object you pass to the slug field.

# Import slugify to generate slugs from strings
from django.utils.text import slugify 

class Album(models.Model):
    artist = models.CharField(max_length=250)
    album_title = models.CharField(max_length=250)
    genre = models.CharField(max_length=100)
    album_logo = models.CharField(max_length=1000,blank=True)
    slug = models.SlugField(unique=True)

    def save(self, *args, **kwargs):
        self.slug = slugify(self.album_title)
        super(Album, self).save(*args, **kwargs)

    def __str__(self):
        return self.album_title    

class Song(models.Model):
    album = models.ForeignKey(Album, on_delete=models.CASCADE)
    artist = models.CharField(max_length=250, blank=True)
    file_type = models.CharField(max_length=10)
    song_title = models.CharField(max_length=100)
    slug = models.SlugField(max_length=100, unique=True)

    def save(self, *args, **kwargs):
        self.slug = slugify(self.song_title)
        super(Song, self).save(*args, **kwargs)    

    def __str__(self):
        return self.artist

How to create a unique slug in Django

I use this snippet for generating unique slug and my typical save method look like below

slug will be Django SlugField with blank=True but enforce slug in save method.

typical save method for Need model might look below

def save(self, **kwargs):
    slug_str = "%s %s" % (self.title, self.us_zip) 
    unique_slugify(self, slug_str) 
    super(Need, self).save(**kwargs)

and this will generate slug like buy-a-new-bike_Boston-MA-02111 , buy-a-new-bike_Boston-MA-02111-1 and so on. Output might be little different but you can always go through snippet and customize to your needs.

How to generate slug based on title of post in django?

I recommend checking out the Django documentation for slugify. You will need to override the save method of your model to do this, so your new code will most likely look something like this:

from django.utils.text import slugify
slug=models.SlugField()

def save(self,*args,**kwargs):
    self.slug=slugify(self.title)
    super(Post,self).save(*args,**kwargs)

I would keep in mind the unique parameter that you can set to either true or false in your slugfield.

How to create a slug from non-English values?

To solve this problem you can use the unidecode and slugify.

you should install it by pip install unidecode

from unidecode import unidecode
from django.template import defaultfilters
slug = defaultfilters.slugify(unidecode(input_text))

example:

import unidecode
a = unidecode.unidecode('привет')
the answer will be ('privet') #this is just an example in russian.
and after that you can apply slugify

Django - How to create a slug url that works?

So guys, I finally figured it out. My slug url is now working perfectly.

The error was actually coming from my list page. The url to the detail view wasn't written well. I didn't include a namespace. So instead of <a href="{{ post.get_absolute_url }}"> it becomes <a href="{% url 'blog:blog_post' post.slug %}">
I also deleted redundant lines of codes in my models and my views page.

Here is what it looks like now.

models.py:

class Post(models.Model):
    STATUS_CHOICES = (
    ('draft', 'Draft'),
    ('published', 'Published'),
)

   title = models.CharField(max_length=200)
   slug = models.SlugField(max_length=200, default="")
   author = models.ForeignKey('auth.User', on_delete=models.CASCADE)
   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=10, choices=STATUS_CHOICES, default='draft')

   class Meta:
     ordering = ('-publish',)

   def __str__(self):
     return self.title

views.py

class BlogHomePageView(ListView):
   model = Post
   template_name = 'blog/index.html'
   context_object_name = 'posts'

class PostDetailView(DetailView):
    model = Post
    template_name = 'blog/post.html'
    context_object_name = 'post'

blog/urls.py:

    urlpatterns = [
      path('', BlogHomePageView.as_view(), name='blog_home'),
      path('post/<slug:slug>/', PostDetailView.as_view(), name='blog_post'),
    ]

list page(index.html):

{% extends 'blog/base.html' %}

    {% block title %}Home{% endblock %}

    <!-- Page Header -->
    {% block page_header %}
        <div class="site-heading">
            <h1>Toluwalemi's Blog</h1>
            <span class="subheading">Official Blog</span>
        </div>
    {% endblock %}

    <!-- Main Content -->
    {% block content %}
        {% for post in posts %}
            <div class="post-preview">
                <a href="{% url 'blog:blog_post' post.slug %}">
                    <h2 class="post-title">
                        {{ post.title }}
                    </h2>

How to Save Slug Automatic in Django?

exclude the slug field from Django Admin by creating a ModelAdmin class as below,

# admin.py

class ArticleAdmin(admin.ModelAdmin):
    exclude = ('slug',)
admin.site.register(Article, ArticleAdmin)

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