从HTML文件中提取信息

Question

我试图从使用Python的大型HTML页面中提取特定信息/链接。例如，从一个IMDB页输出如下HTML，我试图提取电影的链接，这是这样的：

HREF =“/标题/ tt2388771/ref_ = nm_flmg_act_1？”丛林之书：起源

使用下面的Python代码似乎没有工作：

from urllib2 import urlopen 
import re 

source = urlopen("http://www.imdb.com/name/nm0000288/").read() 

print re.findall('href="/title/', source) 
print source 

任何帮助/建议？

<span class="ghost">|</span> <a href="#self" 
 
onclick="handleFilmoJumpto(this);" data-category="self">Self</a></a> 
 
<span class="ghost">|</span> <a href="#archive_footage" 
 
onclick="handleFilmoJumpto(this);" data-category="archive_footage">Archive footage</a></a> 
 
</div> 
 
<div id="filmography"> 
 
<div id="filmo-head-actor" class="head" data-category="actor" onclick="toggleFilmoCategory(this);"> 
 
<span id="hide-actor" class="hide-link" 
 
>Hide&nbsp;<img src="http://ia.media-imdb.com/images/G/01/imdb/images/icons/hide-1061525577._CB358668250_.png" class="absmiddle" alt="Hide" width="18" height="16"></span> 
 
<span id="show-actor" class="show-link" 
 
>Show&nbsp;<img src="http://ia.media-imdb.com/images/G/01/imdb/images/icons/show-582987296._CB358668248_.png" class="absmiddle" alt="Show" width="18" height="16"></span> 
 
<a name="actor">Actor</a> (49 credits) 
 
</div> 
 
<div class="filmo-category-section" 
 
> 
 
<div class="filmo-row odd" id="actor-tt2388771"> 
 
<span class="year_column"> 
 
&nbsp;2017 
 
</span> 
 
<b><a href="/title/tt2388771/?ref_=nm_flmg_act_1" 
 
>Jungle Book: Origins</a></b> 
 
(<a href="/r/legacy-inprod-name/title/tt2388771" class="in_production">filming</a>) 
 
<br/> 
 
<a href="/character/ch0011743/?ref_=nm_flmg_act_1" 
 
>Bagheera</a> 
 
</div> 
 
<div class="filmo-row even" id="actor-tt1596363"> 
 
<span class="year_column"> 
 
&nbsp;2016 
 
</span> 
 
<b><a href="/title/tt1596363/?ref_=nm_flmg_act_2" 
 
>The Big Short</a></b> 
 
(<a href="/r/legacy-inprod-name/title/tt1596363" class="in_production">filming</a>) 
 
<br/> 
 
Michael Burry 
 
</div>

Answer 1

无需使用正则表达式来搜索信息的HTML文件。改为使用the worldwide famous Beautiful Soup。

示例你的使用情况：

from urllib2 import urlopen 
from bs4 import BeautifulSoup 
import re 

source = urlopen("http://www.imdb.com/name/nm0000288/").read() 

soup = BeautifulSoup(source) 
soup.findAll('a', href=re.compile('^/title/'))