如何组合所有3合1 re.findall()??(python 2.7&& Regular Expressions)

时间:2015-02-19 23:15:09

标签: python html regex html-parsing

Filter1=re.findall(r'<span (.*?)</span>',PageSource) 
Filter2=re.findall(r'<a href=.*title="(.*?)" >',PageSource) 
Filter3=re.findall(r'<span class=.*?<b>(.*?)</b>.*?',PageSource)

如何在一行代码中完成...就像这样:

Filter=re.findall(r'  ',PageSource)

我试过这种方式:

Filter=re.findall(r'<span (.*?)</span>'+
                  r'<a href=.*title="(.*?)" >'+
                  r'<span class=.*?<b>(.*?)</b>.*?',PageSource)

但它没有用。

1 个答案:

答案 0 :(得分:2)

如何使用 HTML Parser 呢?

示例,使用BeautifulSoup

from bs4 import BeautifulSoup

data = "your HTML here"
soup = BeautifulSoup(data)

span_texts = [span.text for span in soup.find_all('span')]
a_titles = [a['title'] for a in soup.find_all('a', title=True)]
b_texts = [b.text for b in soup.select('span[class] > b')]

result = span_texts + a_titles + b_texts

演示:

>>> from bs4 import BeautifulSoup
>>> 
>>> data = """
... <div>
...     <span>Span's text</span>
...     <a title="A title">link</a>
...     <span class="test"><b>B's text</b></span>
... </div>
... """
>>> soup = BeautifulSoup(data)
>>> 
>>> span_texts = [span.text for span in soup.find_all('span')]
>>> a_titles = [a['title'] for a in soup.find_all('a', title=True)]
>>> b_texts = [b.text for b in soup.select('span[class] > b')]
>>> 
>>> result = span_texts + a_titles + b_texts
>>> print result
[u"Span's text", u"B's text", 'A title', u"B's text"]

除此之外,你的正则表达式非常不同并且用于不同的目的 - 我不会尝试挤压不可压缩的,将它们分开并将结果合并到一个列表中。