Put a button in the creation of a model - django

Question

I would like to make sure that in addition to the title, color and description, I would like to make sure that when I create a model it makes me choose which url to render it when Ia person clicks the button. It's possible?
models.py
from django.db import models
from colorfield.fields import ColorField
class Aziende(models.Model):
immagine = models.ImageField()
nome = models.CharField(max_length=250)
prezzo = models.FloatField()
descrizione = models.CharField(max_length=250)
nome_color = ColorField(default='#FF0000')
def __str__(self):
return self.nome
class Meta:
verbose_name_plural = "Aziende"
home.html
[...]
<div class="container">
<div class="row">
{% for Aziende in Borsa %}
<div class="col"><br>
<div class="container text-center">
<div class="card" style="width: 18rem;">
<img src="{{ Aziende.immagine.url }}" class="card-img-top" alt="...">
<div class="card-body">
<h5 class="title" style="color: {{Aziende.nome_color}}" >{{Aziende.nome}}</h5>
<p class="card-text">{{Aziende.descrizione}}</p>
<p class="card-text fst-italic text-primary">Valore Attuale: €{{Aziende.prezzo}}</p>
Più info
</div>
</div>
</div>
</div>
{% endfor %}
</div>
</div>
[...]
urls.py (1)
[..]
urlpatterns = [
path('admin/', admin.site.urls),
path('Borsa/', include('Borsa.urls')),
path('accounts/', include('django.contrib.auth.urls')),
path('', TemplateView.as_view(template_name='home.html'), name='home'),
]+static(settings.MEDIA_URL, document_root =settings.MEDIA_ROOT)
urls.py (2)
[...]
urlpatterns = [
path('', views.Borsa),
path('Investimenti', views.Investimenti, name='Investimenti'),
]
Thanks in advance!

Answer 1

You can do this way
<div class="dropdown">
<button class="dropbtn">Dropdown</button>
<div class="dropdown-content">
Home
admin
Borsa
Link 3
</div>
</div>