Python :Page Navigator Maximum Value Scraper - 只获取最后一个值的输出

Question

这是我创建的程序，用于从 list.I 的每个类别部分中提取最大页面值我无法获取所有值，我只是获取中最后一个值的值list.What 我需要进行更改才能获得所有输出。

import bs4
from urllib.request import urlopen as uReq
from bs4 import BeautifulSoup as soup

#List for extended links to the base url

links = ['Link_1/','Link_2/','Link_3/']
#Function to find out the biggest number present in the page navigation
#section.Every element before 'Next→' is consist of the upper limit

def page_no():
    bs = soup(page_html, "html.parser")
    max_page = bs.find('a',{'class':'next page-numbers'}).findPrevious().text
   print(max_page)

#url loop
for url in links:
    my_urls ='http://example.com/category/{}/'.format(url)

# opening up connection,grabbing the page
uClient = uReq(my_urls)
page_html = uClient.read()
uClient.close()
page_no()

页面导航器示例： 1 2 3 … 15 Next →

提前致谢

Answer 1

您需要将 page_html 放在函数内部并缩进最后 4 行。此外，最好 return max_page 值，以便您可以在函数外使用它。

def page_no(page_html): 
    bs = soup(page_html, "html.parser")
    max_page = bs.find('a',{'class':'next page-numbers'}).findPrevious().text
    return max_page

#url loop 
for url in links: 
    my_urls='http://example.com/category/{}/'.format(url) 
    # opening up connection,grabbing the page 
    uClient = uReq(my_urls) 
    page_html = uClient.read()
    uClient.close() 
    max_page = page_no(page_html)
    print(max_page)

Python :Page Navigator Maximum Value Scraper - 只获取最后一个值的输出

Python :Page Navigator Maximum Value Scraper - Only getting the output of last value

python-3.x

beautifulsoup

urllib