1

我想下载一些 html 页面并提取信息，每个 HTML 页面都有这个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 name ( Jhone) 、family ( Doe) 和 tel( 33727464)，我使用beausiful soup通过 id 访问这些 span 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__()

但我不知道如何将数据提取到这些标签中。我尝试使用children和content属性，但是当我使用主题作为tag它返回时None：

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

但我收到此错误：

'NoneType' object has no attribute 'children'

而当我在它上面使用str () 时，它不是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)
4

2 回答 2

3

我找到了几种方法来做到这一点。

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 :)

告诉我它是否有帮助。

于 2012-07-28T15:13:42.447 回答
1

如果您熟悉 xpath，请改用 lxml 和 etree：

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]
于 2012-07-28T21:22:01.867 回答