What is the best, cleanest and shortest way to check if a given value is valid URL when working with models?

Question

I rather want to use Django's built-in functionalities as much as possible and avoid implementing stuff myself as much as possible!

Why doesn't the following code issue exceptions when given a non-URL value?

models.py:

from django.core.validators import URLValidator
from django.db import models

class Snapshot(models.Model):
    url = models.URLField(validators=[URLValidator])

views.py:

from django.http import HttpResponse
from .models import Snapshot

def index(request):
    a = Snapshot(url='gott ist tot')
    a.save()

Answer 1

Because this validator is run when you use a django form. More information about validators on the doc : https://docs.djangoproject.com/en/4.1/ref/validators/

if you do a form :

from django import forms
from .models import Snapshot

class SnshotForm(forms.ModelForm):
    class Meta:
        model = Snapshot
        fields = ('url', )

and your views.py :

from django.http import HttpResponse
from .forms import SnapshotForm

def index(request):
    a = SnapshotForm(data={'url': 'gott ist tot'})
    if a .is_valid()
        a.save()
    else:
        print(a.errors)

Your validator will be run and you will see the errors form message

Without using form, you can call manually validator in your view or in the save method:

# in views
from django.core.validators import URLValidator
from django.core.exceptions import ValidationError

def index(request):
    url_validator = URLValidator()
    url = 'gott ist tot'

    is_valid_url = False
    try:
        url_validator(url)
    except ValidationError:
       pass

    if is_valid_url:
        a = Snapshot(url=url)
        a.save()
    else:
        print(a.errors)

Be careful ! I do not recommanded to bypass the validator with forms, i think it is the better way for maximizing usage of django builtins funtions