How to write URL and VIEWS - django

Question

Below is my model
class Movie(models.Model):
name = models.CharField(max_length=55)
artists = models.ManyToManyField(Artist)
def __str__(self):
return self.name
Below is the View:
def moviesView(request):
movies = Movie.objects.all()
context = {
"movies": movies
}
return render(request, 'movies/movies.html', context=context)
def movieView(request, name):
print(name)
movie = Movie.object.get(name=name)
context = {
"movie": movie
}
return render(request, 'movies/movie.html', context=context)
Below are the URLS:
urlpatterns = [
path('movies', moviesView, name='movies'),
re_path('movies/(\d+)', movieView, name='movie'),
path('artists', artistView, name='artists')
]
Below is the template:
<h1>Movies</h1>
{% for movie in movies %}
{{ movie.name }}
<h3>{{ movie.name }}</h3>
{% for artist in movie.artists.all %}
<ul>
<li>{{ artist.name }}</li>
</ul>
{% endfor %}
{% endfor %}
If I click a movie avengers, it should carry to another page with movie details of avengers as mentioned in the model:
I need to frame the url as: http://127.0.0.1:8000/movies/avengers

Answer 1

You did not specify a parameter. But even if you did, it would not accept it, since \d+ only accepts a sequence of digits, not strings.
You can work with:
urlpatterns = [
path('movies/', moviesView, name='movies'),
path('movies/(?P<name>.*)/', movieView, name='movie'),
path('artists/', artistView, name='artists')
]
but it is likely easier to work with path instead:
urlpatterns = [
path('movies/', moviesView, name='movies'),
re_path('movies/<str:name>/', movieView, name='movie'),
path('artists/', artistView, name='artists')
]
It is also not a good idea to work with a name. A lot of characters will be percent encoded [wiki] resulting in ugly URLs. Django uses slugs to make visually pleasant URLs.