print(len(qualified))
покажет длину полного списка вашей ошибки. Вы можете получить количество ссылок в каждой, добавив i = 0
после while pg<=number_of_pages:
и i += 1
после if n>2:
, а затем добавив print(i)
до или после pg +=1
. Тогда код будет таким:
from bs4 import BeautifulSoup as bs
import requests
import pandas
pg = 1
url ="https://stackoverflow.com/search?page="+str(pg)+"&tab=Relevance&q=scrappy%20python"
src = requests.get(url).text
soup = bs(src,'html.parser')
pages = soup.findAll('a',{'class' : 's-pagination--item js-pagination-item'})
number_of_pages = len(pages)
print(number_of_pages)
qualified=[]
while pg<=number_of_pages:
i = 0
print("In Page :"+str(pg))
url = "https://stackoverflow.com/search?page=" + str(pg) + "&tab=Relevance&q=scrappy%20python"
src = requests.get(url).text
soup = bs(src, 'html.parser')
a_links = soup.findAll('a',{'class':'question-hyperlink'})
span_links = soup.findAll('span',{'class':'vote-count-post'})
hrefs = []
for a_link in a_links:
hrefs.append(a_link.get('href'))
for link in range(len(span_links)):
vote = span_links[link].strong.text
n = int(vote)
if n>2:
i += 1
the_link = 'https://stackoverflow.com' + hrefs[link]
qualified.append(the_link)
print(i)
pg +=1
#print(qualified)
Вывод:
6
In Page :1
1
In Page :2
4
In Page :3
2
In Page :4
3
In Page :5
2
In Page :6
2