我想从this page获取链接并将它们放在列表中。
我有以下代码：
import bs4 as bs
import urllib.request
source = urllib.request.urlopen('http://www.gcoins.net/en/catalog/236').read()
soup = bs.BeautifulSoup(source,'lxml')
links = soup.find_all('a', attrs={'class': 'view'})
print(links)
它产生以下输出：
[<a class="view" href="/en/catalog/view/514">
<img alt="View details" height="32" src="/img/actions/file.png" title="View details" width="32"/>
</a>,
"""There are 28 lines more"""
<a class="view" href="/en/catalog/view/565">
<img alt="View details" height="32" src="/img/actions/file.png" title="View details" width="32"/>
</a>]
我需要得到以下信息：
[/en/catalog/view/514, ... , '/en/catalog/view/565']但是然后我继续添加以下内容：
href_value = links.get('href')我遇到了错误。 最佳答案
尝试：
soup = bs.BeautifulSoup(source,'lxml')
links = [i.get("href") for i in soup.find_all('a', attrs={'class': 'view'})]
print(links)
输出：
['/en/catalog/view/514', '/en/catalog/view/515', '/en/catalog/view/179080', '/en/catalog/view/45518', '/en/catalog/view/521', '/en/catalog/view/111429', '/en/catalog/view/522', '/en/catalog/view/182223', '/en/catalog/view/168153', '/en/catalog/view/523', '/en/catalog/view/524', '/en/catalog/view/60228', '/en/catalog/view/525', '/en/catalog/view/539', '/en/catalog/view/540', '/en/catalog/view/31642', '/en/catalog/view/553', '/en/catalog/view/558', '/en/catalog/view/559', '/en/catalog/view/77672', '/en/catalog/view/560', '/en/catalog/view/55377', '/en/catalog/view/55379', '/en/catalog/view/32001', '/en/catalog/view/561', '/en/catalog/view/562', '/en/catalog/view/72185', '/en/catalog/view/563', '/en/catalog/view/564', '/en/catalog/view/565']
关于python - 使用BeautifulSoup将所有href刮到列表中，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/50706832/