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I wrote the script:

 import requests import urllib.request from bs4 import BeautifulSoup r = requests.get('http://mytricolortv.com/') s1 = BeautifulSoup(r.text,"html.parser") m = s1.find_all('script') print(m)

Can't get to src. I need to extend the path //pagead2.googlesyndication.com/pagead/js/adsbygoogle.js

Help me please.

  • Thank you, Axenow! - Ivan Petrov

2 answers 2

This is how it will work. So that you do not catch exceptions then that src is not in the soup.

 from bs4 import BeautifulSoup s1 = BeautifulSoup(open("t.html", "r").read() ,"html.parser") m = s1.find_all('script') sources=s1.findAll('script',{"src":True}) for source in sources: print(source['src'])
  • Thank you, Axenow! - Ivan Petrov
  • @ Ivan Could you then mark the answer as correct if it helped you? Thanks in advance. - Axenow

There is an option to access src like this:

 s1 = BeautifulSoup(r.text,"html.parser") print( s1.select("img")[0].attrs["src"] )

However, he can return None