Python，Regex：匹配字符串后提取字符串

Question

我想使用正则表达式来匹配一个模式并提取模式的一部分。

我已经抓取了 HTML 数据，一个说明性的 sn-p 看起来像：

</script>
</li>
<li itemprop="itemListElement" itemscope="" itemtype="http://schema.org/ListItem">
<span class="hide" itemprop="position">1</span>
<div class="result-heading">
<a class="project-icon show-outline" href="/projects/quickfixj/" title="Find out more about QuickFIX/J - Open Source Java FIX Engine">
<img alt="QuickFIX/J - Open Source Java FIX Engine Icon" src="//a.fsdn.com/allura/p/quickfixj/icon?1533295730"/></a>
<div class="result-heading-texts">
<a href="/projects/quickfixj/" itemprop="url" title="Find out more 
<a href="/projects/desmoj/" itemprop="url" title="Find out more about DESMO-J"><h2>DESMO-J</h2></a>
<div class="description">
<p class="description-inner">DESMO-<em>J</em> is a framework for 
<a href="/projects/desmoj/files/stats/timeline" title="Downloads This Week">29 This Week</a>
</strong>
<strong>

find_all('a') 更具代表性的子集突出显示问题：

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HTML 当前存储为 BeautifulSoup 对象，即它已被传递：

html_soup= BeautifulSoup(response.text, 'html.parser')

我想在整个对象中搜索/projects/ 的所有实例，并提取后续斜杠之间的字符串。例如：

from "/projects/quickfixj/" I would like to store "quickfixj".

我最初的想法是使用re.findall() 并尝试匹配(/projects/./)*，但这不起作用。

非常感谢任何帮助。

Answer 1

你已经完成了一半

a='''</script>
</li>
<li itemprop="itemListElement" itemscope="" itemtype="http://schema.org/ListItem">
<span class="hide" itemprop="position">1</span>
<div class="result-heading">
<a class="project-icon show-outline" href="/projects/quickfixj/" title="Find out more about QuickFIX/J - Open Source Java FIX Engine">
<img alt="QuickFIX/J - Open Source Java FIX Engine Icon" src="//a.fsdn.com/allura/p/quickfixj/icon?1533295730"/></a>
<div class="result-heading-texts">
<a href="/projects/quickfixj/" itemprop="url" title="Find out more 
<a href="/projects/desmoj/" itemprop="url" title="Find out more about DESMO-J"><h2>DESMO-J</h2></a>
<div class="description">
<p class="description-inner">DESMO-<em>J</em> is a framework for 
<a href="/projects/desmoj/files/stats/timeline" title="Downloads This Week">29 This Week</a>
</strong>
<strong>'''
from bs4 import BeautifulSoup
soup = BeautifulSoup(a,"html.parser")
for i in soup.find_all('a'):
    print(re.findall('/projects/(\w{1,})/',i.get('href')))

如果您需要独特的项目。将最后几行更改为

from bs4 import BeautifulSoup
soup = BeautifulSoup(a,"html.parser")
project_set=set()
for i in soup.find_all('a'):
    project_set.add(*re.findall('/projects/(\w{1,})/',i.get('href')))

print(project_set) #{u'desmoj', u'quickfixj'}

Answer 2

您可以提取所有链接，然后应用正则表达式：

from bs4 import BeautifulSoup

html = '''</script>
</li>
<li itemprop="itemListElement" itemscope="" itemtype="http://schema.org/ListItem">
<span class="hide" itemprop="position">1</span>
<div class="result-heading">
<a class="project-icon show-outline" href="/projects/quickfixj/" title="Find out more about QuickFIX/J - Open Source Java FIX Engine">
<img alt="QuickFIX/J - Open Source Java FIX Engine Icon" src="//a.fsdn.com/allura/p/quickfixj/icon?1533295730"/></a>
<div class="result-heading-texts">
<a href="/projects/quickfixj/" itemprop="url" title="Find out more 
<a href="/projects/desmoj/" itemprop="url" title="Find out more about DESMO-J"><h2>DESMO-J</h2></a>
<div class="description">
<p class="description-inner">DESMO-<em>J</em> is a framework for 
<a href="/projects/desmoj/files/stats/timeline" title="Downloads This Week">29 This Week</a>
</strong>
<strong>'''

html_soup = BeautifulSoup(html, 'html.parser')

links = [i.get('href') for i in html_soup.find_all('a', href=True)]

产量：

['/projects/quickfixj/', '/projects/quickfixj/', '/projects/desmoj/files/stats/timeline']

然后你可以应用你的正则表达式：

cleaned = [re.findall(r'(?<=projects\/)(.*?)\/', i)[0] for i in links]

产量：

['quickfixj', 'quickfixj', 'desmoj']

Answer 3

这样的正则表达式应该可以解决问题(?<=\/projects\/).+?(?=\/)

会像这样工作

import re
regex = "(?<=\/projects\/).+?(?=\/)"
string = "<a href="/projects/quickfixj/" itemprop="url" title="Find out more...."
matches = re.findall(regex, string)
print(matches)

输出：["quickfixj"]