如何从html树中分离标签

Question

这是我的html树

 <li class="taf"><h3><a href="26eOfferCode%3DGSONESTP-----------" id="pa1">
    Citibank <b>Credit Card</b> - Save over 5% on fuel | Citibank.co.in</a>
   </h3>Get the IndianOil Citibank <b>Card</b>. Apply Now! 
   <br />
   <a href="e%253DGOOGLE ------">Get 10X Rewards On Shopping</a> -
   <a href="S%2526eOfferCode%253DGSCCSLEX ------">Save Over 5% On Fuel</a>
   <br />
   <cite>www.citibank.co.in/<b>CreditCards</b></cite>
</li>

从这个html中，我需要提取<br>标记

第1行：获得IndianOil花旗银行卡。立即申请！

line2：购物可获得10倍奖励-燃油可节省5%以上

在python中应该怎么做？

Answer 1

我想您只是在每次<br/>之前要求行。

下面的代码将通过剥离<b>和<a>标记并打印following-sibling是<br/>的每个元素的.tail来为您提供的示例执行此操作。

from lxml import etree
doc = etree.HTML("""
<li class="taf"><h3><a href="26eOfferCode%3DGSONESTP-----------" id="pa1">
    Citibank <b>Credit Card</b> - Save over 5% on fuel | Citibank.co.in</a>
   </h3>Get the IndianOil Citibank <b>Card</b>. Apply Now! 
   <br />
   <a href="e%253DGOOGLE ------">Get 10X Rewards On Shopping</a> -
   <a href="S%2526eOfferCode%253DGSCCSLEX ------">Save Over 5% On Fuel</a>
   <br />
   <cite>www.citibank.co.in/<b>CreditCards</b></cite>
</li>""")
etree.strip_tags(doc,'a','b')
for element in doc.xpath('//*[following-sibling::*[name()="br"]]'):
  print repr(element.tail.strip())

收益率：

'Get the IndianOil Citibank Card. Apply Now!'
'Get 10X Rewards On Shopping -n   Save Over 5% On Fuel'

Answer 2

与HTML的所有解析一样，您需要对HTML的格式做出一些假设。如果我们可以假设前一行是从<br>标签到块级标签或另一个<br>之前的所有内容，那么我们可以执行以下操作。。。

from BeautifulSoup import BeautifulSoup
doc = """
   <li class="taf"><h3><a href="26eOfferCode%3DGSONESTP-----------" id="pa1">
    Citibank <b>Credit Card</b> - Save over 5% on fuel | Citibank.co.in</a>
    </h3>Get the IndianOil Citibank <b>Card</b>. Apply Now!
    <br />
    <a href="e%253DGOOGLE ------">Get 10X Rewards On Shopping</a> -
    <a href="S%2526eOfferCode%253DGSCCSLEX ------">Save Over 5% On Fuel</a>
    <br />
    <cite>www.citibank.co.in/<b>CreditCards</b></cite>
</li>
"""
soup = BeautifulSoup(doc)

现在我们已经解析了HTML，接下来我们定义了不想作为行的一部分处理的标记列表。确实还有其他块标记，但这对这个HTML来说是有效的。

block_tags = ["div", "p", "h1", "h2", "h3", "h4", "h5", "h6", "br"]

我们循环遍历每个<br>标签，再遍历它的兄弟标签，直到没有更多标签，或者达到块级标签。每次循环时，我们都会将节点添加到行的前面。NavigableStrings没有name属性，但我们希望将它们包括在内，因此while循环中有两部分测试。

for node in soup.findAll("br"):
    line = ""
    sibling = node.previousSibling
    while sibling is not None and (not hasattr(sibling, "name") or sibling.name not in block_tags):
        line = unicode(sibling) + line
        sibling = sibling.previousSibling
    print line

Answer 3

不中继<br>标签的解决方案：

import lxml.html
html = "..."
tree = lxml.html.fromstring(html)
line1 = ''.join(tree.xpath('//li[@class="taf"]/text() | b/text()')[:3]).strip()
line2 = ' - '.join(tree.xpath('//li[@class="taf"]//a[not(@id)]/text()'))

Answer 4

我不知道你是想用lxml还是想用漂亮的汤。但是对于使用xpath的lxml，这里有一个示例

import lxml
from lxml import etree
import urllib2
response = urllib2.urlopen('your url here')
html = response.read()
imdb = etree.HTML(html)
titles = imdb.xpath('/html/body/li/a/text()')//xpath for "line 2" data.[use firebug]

我使用的xpath用于给定的html片段。它可能会在原始上下文中发生变化。

你也可以试试cssselect in lxml。

import lxml.html
import urllib
data = urllib.urlopen('your url').read() 
doc = lxml.html.fromstring(data)
elements = doc.cssselect('your csspath here') // CSSpath[using firebug extension]
for element in elements:
      print element.text_content()