使用Beautifulsoup从网站上剥离链接

Question

我在Python中使用beautifulsoup时，努力从下面的“结果”中获取链接（即/d/Hinchinbrook+25691+Masjid-Bilal）。请帮帮忙？

结果：

<div class="subtitleLink"><a href="/d/Hinchinbrook+25691+Masjid-Bilal"><b>Masjid Bilal</b></a></div>

代码：

url1 = "http://www.salatomatic.com/c/Sydney+168"
content1 = urllib2.urlopen(url1).read()
soup = BeautifulSoup(content1)
results = soup.findAll("div", {"class" : "subtitleLink"})
for result in results :
print result
br = result.find('a')
pos = br.get_text()
print pos

Answer 1

import urllib2
from bs4 import BeautifulSoup

url1 = "http://www.salatomatic.com/c/Sydney+168"
content1 = urllib2.urlopen(url1).read()
soup = BeautifulSoup(content1)
for link in soup.findAll('a'):
   print link.get('href')

如果您想要所有链接，这应该有用。如果没有，请告诉我。

Answer 2

get_text方法仅返回标记的字符串组件。要获取此处的链接，请将其引用为attribute。对于此特定情况，您可以将br.get_text()更改为br['href']以获得所需的结果。

...
>>> br = result.find('a')
>>> pos = br['href']
>>> print pos
/d/Hinchinbrook+25691+Masjid-Bilal