Python 仅获取URL组
我有以下汤Python 仅获取URL组,python,python-3.x,beautifulsoup,python-requests,Python,Python 3.x,Beautifulsoup,Python Requests,我有以下汤 <a href="https://www.abc1.com"> <h3>ABC1</h3> </a> <a href="https://www.abc2.com"> <h3>ABC2</h3> </a> <a href="https://www.abc3.com"> <h3>ABC3</h3> </a> 但这显示的是空
<a href="https://www.abc1.com">
<h3>ABC1</h3>
</a>
<a href="https://www.abc2.com">
<h3>ABC2</h3>
</a>
<a href="https://www.abc3.com">
<h3>ABC3</h3>
</a>
但这显示的是空数组,
像这样,
[][][]
有谁知道更好的方法吗?我可以使用以下代码获取href:-
for link in links:
print(link['href'])
什么是链接和标题?添加完整的代码这是一种汤,实际上我是根据自己的喜好命名的。@Uditharavisht请再看一次这个问题
for link in links:
print(link['href'])
cont = soup.find_all('a')
link = []
for href in cont:
print(link.append(href.get('href')))
#o/p
link
['https://www.abc1.com', 'https://www.abc2.com', 'https://www.abc3.com']