Extract Title Tags BeautifulSoup - beautifulsoup

Question

I need help because I wanted to write code for finding out the title tags on a website. Although I used the code from another question and applied it to this scenario, there are no title tags whenever I print 'Beschreibung'.
from bs4 import BeautifulSoup
import requests
import pandas as pd
import urllib.parse
webseite = 'https://www.entega.de/sitemap/'
response = requests.get(webseite)
response.status_code
soup = BeautifulSoup (response.content, 'html.parser')
result_container = soup.find_all('div', {'class':'clearfix'})
url_part_1 = 'https://www.entega.de/sitemap/'
url_part_2 = []
for item in result_container:
for link in item.find_all ('a', {'class':'modSitemap__lvl1Link ui-link' }):
url_part_2.append (link.get ('href'))
url_joined = []
for i in url_part_2:
url_joined.append (urllib.parse.urljoin(url_part_1,i))
Überschrift = []
Beschreibung = []
Verlinkungen = []
for link in url_joined:
response = requests.get (link)
soup = BeautifulSoup (response.content, 'html.parser')
Beschreibung.append(soup.find_all('a', title=True, class_='modSitemap__lvl1Link ui-link'))

Answer 1

You are getting nothing because these links don't have an <a class="modSitemap__lvl1Link ui-link"> tag. They do have classes that start with that string though. You could expand to that. Or you can simply just get the <a> tags that have a title attribute.
So change your loop to either:
import re
for link in url_joined:
response = requests.get (link)
soup = BeautifulSoup (response.content, 'html.parser')
Beschreibung.append(soup.find_all('a', {'class':re.compile("^modSitemap__lvl1Link ui-link")}, title=True, ))
or
for link in url_joined:
response = requests.get (link)
soup = BeautifulSoup (response.content, 'html.parser')
Beschreibung.append(soup.find_all('a', title=True))