如何在Python中提取网页的某些部分

Question

目标网页:http: //www.immi.gov.au/skilled/general-skilled-migration/estimated-allocation-times.htm

我要提取的部分:

  <tr>
  <td>Skilled &ndash; Independent (Residence) subclass 885<br />online</td>
  <td>N/A</td>
  <td>N/A</td>
  <td>N/A</td>
  <td>15 May 2011</td>
  <td>N/A</td>
  </tr>

一旦代码通过搜索关键字" 子类885
在线 " 找到此部分,它应该打印第5个标签内的日期,即" 2011年5月15日 ",如上所示.

它只是一个监视我自己的监视我的移民申请的进展.

Answer 1

您可能希望将此作为起点:

Python 2.6.7 (r267:88850, Jun 13 2011, 22:03:32) 
[GCC 4.6.1 20110608 (prerelease)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import urllib2, re
>>> from BeautifulSoup import BeautifulSoup
>>> urllib2.urlopen('http://www.immi.gov.au/skilled/general-skilled-migration/estimated-allocation-times.htm')
<addinfourl at 139158380 whose fp = <socket._fileobject object at 0x84aa2ac>>
>>> html = _.read()
>>> soup = BeautifulSoup(html)
>>> soup.find(text = re.compile('\\bsubclass 885\\b')).parent.parent.find('td', text = re.compile(' [0-9]{4}$'))
u'15 May 2011'

Answer 2

" Beau - ootiful Soo - oop!

Beau - ootiful Soo - oop!

e - e - 晚上的soo - oop,

美丽,美丽 - FUL SOUP!"

- 刘易斯卡罗尔,爱丽丝梦游仙境

我认为这正是他的想法!

模拟海龟可能会做这样的事情:

>>> from BeautifulSoup import BeautifulSoup
>>> import urllib2
>>> url = 'http://www.immi.gov.au/skilled/general-skilled-migration/estimated-allocation-times.htm'
>>> page = urllib2.urlopen(url)
>>> soup = BeautifulSoup(page)
>>> for row in soup.html.body.findAll('tr'):
...     data = row.findAll('td')
...     if data and 'subclass 885online' in data[0].text:
...         print data[4].text
... 
15 May 2011

但我不确定它会有所帮助,因为那个日期已经过去了!

祝你好运!