BeautifulSoup not working?

Question

Python

Actually, I want to get the address of the picture. But the problem is that nothing works. Doesn't even show the document.

import urllib.request as urllib2
from bs4 import BeautifulSoup
page = urllib2.urlopen("http://www.vokrugsveta.ru/vs/").read()
soup = BeautifulSoup(page)
print(soup.find("img", class='framepic'))

Answer 1

import urllib.request as urllib2
from bs4 import BeautifulSoup
page = urllib2.urlopen("http://www.vokrugsveta.ru/vs/").read()
soup = BeautifulSoup(page)
imglink = soup.find("img", {"class": "framepic"}).get('src')
print (imglink)

boys use lxml instead of soup

Answer 2

class - a reserved word of the language in Bs4 instead of it in the parameters, you can useclass_

# Варианты:
print(soup.find("img", class_='framepic'))
print(soup.find("img", {'class':'framepic'}))
print(soup.find("img", 'framepic'))

Answer 3

print page покажет содержимое?