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python - 使用 BeautifulSoup 将所有 href 刮到列表中

问题描述

我想从这个页面抓取链接并将它们放在一个列表中。

我有这个代码：

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']

但后来我继续添加以下内容：href_value = links.get('href')我遇到了一个错误。

标签： pythonlistweb-scrapingbeautifulsoup

解决方案

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']

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