Question 43127 in Html Question posted in
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How do I parse site html with "span"?

I’am new in web-parsing and I ran into a problem.
I wanna extract number of watching from the site:
enter image description here

I wrote this code for that purpose:

url = 'https://www.kommersant.ru/doc/4638344'
request = requests.get(url)
req = request.text
soup=BeautifulSoup(req, 'html')
abcd = soup.find("div", class_="doc_sharing__body js-social").find("span", class_="sharing")
print(abcd)

But result of this query is "None"

What’s the problem? Please, help!

I try many iterations with different variation of my code (for example):

url = 'https://www.kommersant.ru/doc/4638344'
request = requests.get(url)
req = request.text
soup=BeautifulSoup(req, 'html')
abcd =  soup.find_all("span", attr = {'class':'sharing'}
print(abcd)

But I have same result!

Tags:

2

Answers


  1. 0

    I believe that something like this script can help you retrieve the number of views (I am assuming that this number is the one inside the title tag):

    import requests
from bs4 import BeautifulSoup

url = "https://www.kommersant.ru/doc/4638344"
response = requests.get(url)
soup = BeautifulSoup(response.content, "html.parser")

sharing_div = soup.find("div", class_="doc_sharing__body")

title = sharing_div.find("span", class_="sharing")["title"]

watching_number = title.split(": ")[1]

print(watching_number)

    Note: Maybe you’ll have to use .find_all(), instead of .find(). In this case, you have to inspect the retrieved results, to check which of them contains the desired title tag.

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  2. 0

    The view count is stored in the data-article-views= parameter:

    import requests
from bs4 import BeautifulSoup

url = 'https://www.kommersant.ru/doc/4638344'
soup = BeautifulSoup(requests.get(url).content, 'html.parser')

for article in soup.select('article[data-article-views]'):
    print(article['data-article-title'], article['data-article-views'])

    Prints:

    «Газпром» начал поставлять газ в Сербию по «Турецкому потоку» 14541
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