how to define parser when using BS4 in python - python-3.x

Question

#!/usr/bin/env python
import requests
from bs4 import BeautifulSoup
url = "https://www.youtube.com/channel/UCaKt8dvEIPnEHWSbLYhzrxg/videos"
response = requests.get(url)
# parse html
page = str(BeautifulSoup(response.content))
def getURL(page):
"""
:param page: html of web page (here: Python home page)
:return: urls in that page
"""
start_link = page.find("a href")
if start_link == -1:
return None, 0
start_quote = page.find('"', start_link)
end_quote = page.find('"', start_quote + 1)
url = page[start_quote + 1: end_quote]
return url, end_quote
while True:
url, n = getURL(page)
page = page[n:]
if url:
print(url)
else:
break
I am using above code to get list of all youtube videos on webpage. If i try to do this. I get following error
The code that caused this warning is on line 9 of the file C:/Users/PycharmProjects/ReadCSVFile/venv/Links.py. To get rid of this warning, change code that looks like this:
I did and started using html but some different error came .
I am using Python 3.0 . I am using IDE Pycharm.
Can someone please help me this.

Answer 1

its not error, but warning you didn't set parser which can be 'html.parser', 'lxml', 'xml'. change it to like
page = BeautifulSoup(response.content, 'html.parser')
your code above actually not doing what BeautifulSoup do, but here the example using it.
#!/usr/bin/env python
import requests
from bs4 import BeautifulSoup
def getURL(url):
"""
:param url: url of web page
:return: urls in that page
"""
response = requests.get(url)
# parse html
page = BeautifulSoup(response.content, 'html.parser')
link_tags = page.find_all('a')
urls = [x.get('href') for x in link_tags]
return urls
url = "https://www.youtube.com/channel/UCaKt8dvEIPnEHWSbLYhzrxg/videos"
all_url = getURL(url)
print('\n'.join(all_url))