PYTHON Lib:bs4 |需要帮助获得

Question

我不知道如何在我的代码上获得不同的URL值。 你能帮我解决这个问题吗

Source HTML

代码：

import bs4 as bs
import urllib.request

sauce = urllib.request.urlopen('https://cointelegraph.com/tags/bitcoin').read()
soup = bs.BeautifulSoup(sauce, 'lxml')

for div in soup.find_all('div', class_='post-preview-item-inline'):
    title = div.find(class_='post-preview-item-inline__title')
    print(title.text)
    body = div.find(class_='post-preview-item-inline__text')
    print(body.text)
    for link in div.select('a'):
        print(link.get('href'))

结果：

Why Bitcoin Price Is Headed to 7-Month Low If $7K Support Fails 
 Bitcoin price continues to yield to bears and could collapse to $6.5K or below if $7,080 and $6,800 fail to hold as support. 
https://cointelegraph.com/news/why-bitcoin-price-is-headed-to-7-month-low-if-7k-support-fails
https://cointelegraph.com/news/why-bitcoin-price-is-headed-to-7-month-low-if-7k-support-fails
javascript:void(0);

目标： 我只想显示一个网址，而不是2显示在结果上面。 我也想知道如何删除javascript:void(0);

谢谢

Answer 1

您正在使用.select，它允许css选择器。就像在css中一样，您可以通过以下方式选择具有适当类的特定链接：

for link in div.select('.post-preview-item-inline__figure-link'):
    print(link.get('href'))