@H_404_8@
我需要得到< a href =>具有类article-additional-info的所有div的值
我是BeautifulSoup的新手
所以我需要网址
"http://www.thehindu.com/news/national/gangrape-case-two-lawyers-claim-to-be-engaged-by-accused/article4332680.ece"
"http://www.thehindu.com/news/cities/Delhi/power-discoms-demand-yet-another-hike-in-charges/article4331482.ece"
@H_404_8@
实现这一目标的最佳方法是什么？
最佳答案
根据您的标准,它返回三个URL(而不是两个) – 您想要过滤掉第三个吗？
基本思想是迭代HTML,只抽取你的类中的那些元素,然后迭代该类中的所有链接,拉出实际的链接：
In [1]: from bs4 import BeautifulSoup
In [2]: html = # your HTML
In [3]: soup = BeautifulSoup(html)
In [4]: for item in soup.find_all(attrs={'class': 'article-additional-info'}):
...: for link in item.find_all('a'):
...: print link.get('href')
...:
http://www.thehindu.com/news/national/gangrape-case-two-lawyers-claim-to-be-engaged-by-accused/article4332680.ece
http://www.thehindu.com/news/cities/Delhi/power-discoms-demand-yet-another-hike-in-charges/article4331482.ece
http://www.thehindu.com/news/cities/Delhi/power-discoms-demand-yet-another-hike-in-charges/article4331482.ece#comments
@H_404_8@
这会将您的搜索范围限制为仅包含article-additional-info类标记的元素,并在其中查找所有锚点(a)标记并获取其相应的href链接.
原文链接：https://www.f2er.com/python/439370.html