Vfrdtyky

#!/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.

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))