Question

我想下载一些HTML页面和提取信息，每个HTML页有这样table tag：

<table class="sobi2Details" style='background-image: url(http://www.imd.ir/components/com_sobi2/images/backgrounds/grey.gif);border-style: solid; border-color: #808080' > 
    <tr> 
     <td><h1>Dr Jhon Doe</h1></td> 
    </tr> 
    <tr> 
     <td></td> 
    </tr> 
    <tr> 
     <td></td> 
    </tr> 
    <tr> 
     <td> 
      <div id="sobi2outer"> 
      <br/> 
      <span id="sobi2Details_field_name" ><span id="sobi2Listing_field_name_label">name:</span>Jhon</span><br/> 
      <span id="sobi2Details_field_family" ><span id="sobi2Listing_field_family_label">family:</span> Doe</span><br/> 
      <span id="sobi2Details_field_tel1" ><span id="sobi2Listing_field_tel1_label">tel:</span> 33727464</span><br/> 
      </div> 
     </td> 
    </tr> 
</table> 

我想访问的域名（Jhone），家庭（Doe）和电话（33727464 ），我用beausiful soup通过ID来访问这些跨度标签：

name=soup.find(id="sobi2Details_field_name").__str__() 
family=soup.find(id="sobi2Details_field_family").__str__() 
tel=soup.find(id="sobi2Details_field_tel1").__str__() 

，但我不知道如何提取数据到这些tags.I tryed使用children和content个属性，但是当我使用的主题为tag它返回None：

name=soup.find(id="sobi2Details_field_name") 
for child in name.children: 
    #process content inside 

，但我得到这个错误：当我使用它STR（）

'NoneType' object has no attribute 'children' 

同时，它不None !! 任何想法？

编辑：我的最终解决方案

soup = BeautifulSoup(page,from_encoding="utf-8") 
name_span=soup.find(id="sobi2Details_field_name").__str__() 
name=name_span.split(':')[-1] 
result = re.sub('</span>', '',name) 

Answer 1

我发现一对夫妇的方式来做到这一点。

from bs4 import BeautifulSoup 
soup = BeautifulSoup(open(path_to_html_file)) 

name_span = soup.find(id="sobi2Details_field_name") 

# First way: split text over ':' 
# This only works because there's always a ':' before the target field 
name = name_span.text.split(':')[1] 

# Second way: iterate over the span strings 
# The element you look for is always the last one 
name = list(name_span.strings)[-1] 

# Third way: iterate over 'next' elements 
name = name_span.next.next.next # you can create a function to do that, it looks ugly :) 

告诉我，如果有帮助。

Answer 2

如果您熟悉使用XPath使用LXML与etree代替：

import urllib2 
from lxml import etree 

opener = urllib2.build_opener() 
root = etree.HTML(opener.open("myUrl").read()) 

print root.xpath("//span[@id='sobi2Details_field_name']/text()")[0]