Acquire model instance as view parameter instead of model id - django

Question

In laravel, it's possible to have access to entire models with type-hinting the parameter with a model class. for example:
routes/web.php
Route::get('posts/{post}', function(\App\Post $post) {
/*
$post is now the model with pk in url and
if the pk is wrong it throws an 404 automatically
*/
});
How is that possible in django? (with functions as view)
def posts(request, post_id):
post = get_model_or_404(Post, pk=post_id)
# ...
The second line is completely boilerplate and so much repeated in my code.

Answer 1

You can make a custom path converter [Django-doc] for that. We can for example make model path converter:
# app/converters.py
from django.shortcuts import get_object_or_404
from django.urls.converters import IntConverter
class ModelConverter(IntConverter):
model = None
def to_python(self, value):
return get_object_or_404(self.model, pk=super().to_python(value))
def to_url(self, value):
if instanceof(value, int):
return str(value)
return str(value.pk)
def model_converter(model):
class Converter(ModelConverter):
model = model
return Converter
Then in your urls.py you can make model converters and register these once:
# app/urls.py
from app.converters import model_converter
from app import views
from app.models import Post
from django.urls import path, register_converter
register_converter(model_converter(Post), 'post')
urlpatterns = [
path('posts/<post:post>', views.post, name='post'),
# …
]
Then in the view, you obtain the Post object (given a post for that primary key exists) through the post parameter:
app/views.py
def post(request, post):
# post is here a Post object
# …
In your template, you can then just pass the Post object in the {% url … %} template tag, so:
<a href="{% url 'post' mypostobject %}">