Question

这是我创建的程序，用于从列表中的每个类别部分中提取最大页面值。我无法获取所有值，我只是获取列表中最后一个值的值。我需要制作才能获得所有输出。

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 →

先谢谢

Answer 1

你需要将page_html放在函数中并缩进最后4行。此外，最好返回max_page值，以便可以使用它来执行函数。

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Python：Page Navigator最大值Scrapper - 仅获取最后一个值的输出

1 个答案: