How to display a link to another site using django?

Question

Django

What are the tools to display another site to the user?
Let's say a user visits my django site, enters a link to another site, and my site displays it in the browser window.

Works with simple sites.

from django.shortcuts import render
from urllib.request import urlopen

def website_index(request):
    title = 'Another website viewer'
    link = 'youtube.com'
    if 'http' not in link:
        link = 'http://' + link

    html_code = str(urlopen(link).read(),'utf-8')

    context = {
        'title': title,
        'html_code': html_code,
    }
    return render(request, 'my_app/index.html', context)
   # потом в index.html
   # {{ html_code|safe }}
   # топорное решение