BeautifulSoup 的 find 方法返回 None 而不是链接

Question

感谢您在这里查看我的问题，我正在尝试从旧的 Reddit 博客页面获取下一页链接 但不知何故 find 方法返回我 None 对象，代码：

 def crawl(self):
        curr_page_url = self.start_url
        curr_page = requests.get(curr_page_url)
        bs = BeautifulSoup(curr_page.text,'lxml')
        # all_links = GetAllLinks(self.start_url)
        nxtlink = bs.find('a',attrs={'rel':'nofollow next'})['href']
        print(nxtlink)

并且 HTML 页面链接是此页面上的 Old Reddit page link 我正在尝试获取下一页的链接 在一个跨度标签中：

<span class="next-button">
    <a href="https://old.reddit.com/r/learnprogramming/?count=25&amp;after=t3_j54ae2" rel="nofollow 
    next">next ›
    </a>
</span>

Answer 1

我认为你必须在请求中添加标头，否则服务器认为你是机器人，这是对的。

试试这个：

import requests
from bs4 import BeautifulSoup

headers = {
    "Accept": "text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8",
    "Accept-Encoding": "gzip, deflate, br",
    "Accept-Language": "en-GB,en;q=0.5",
    "User-Agent": "Mozilla/5.0 (Macintosh; Intel Mac OS X 10.15; rv:81.0) Gecko/20100101 Firefox/81.0",
}

response = requests.get("https://old.reddit.com/r/learnprogramming/", headers=headers).text
soup = BeautifulSoup(response, "html.parser").find('a', attrs={'rel': 'nofollow next'})['href']
print(soup)

输出：

https://old.reddit.com/r/learnprogramming/?count=25&after=t3_j5ezm8