I recently started learning Django. I want to display one news item, but when I open on the link I get an error message:
Cannot resolve keyword 'slug' into field. Choices are: NewsTitles, NewsContent, NewsSlug
Request Method: GET
Request URL: http://127.0.0.1:8000/news/nam-gravida-purus-non/
Django Version: 4.0
Exception Type: FieldError
views.py
from django.views.generic import DetailView
from .models import News
class GetNews(DetailView):
model = News
slug_url_kwarg = 'NewsSlug'
template_name = 'news/single_news.html'
context_object_name = 'single_news'
allow_empty = False
urls.py
from django.urls import path
from .views import GetNews
urlpatterns = [
path('news/<str:NewsSlug>/', GetNews.as_view(), name='news'),
]
models.py
from django.db import models
from django.urls import reverse_lazy
class News(models.Model):
NewsTitles = models.CharField(max_length=120)
NewsContent = models.TextField(max_length=255)
NewsSlug = models.SlugField(max_length=255)
def __str__(self):
return self.NewsTitles
def get_absolute_url(self):
return reverse_lazy('news', kwargs={'NewsSlug': self.NewsSlug})
What am I doing wrong?
CodePudding user response:
First of all do not call your slug "NewSlug" with uppercase but all lower case "newslug" or even better "new_slug", the name itself should be more descriptive as well.
Finally you need to tell your view which field to use, you can define that with the following attribute :
slug_field = "NewSlug"
Note : Attribute of a class should not be camel case but snake case