擅长:python、mysql、java
<p>是的,这是可能的。您可以找到所有的网址,然后提取他们使用反向引用。您可以阅读有关反向引用<a href="http://www.regular-expressions.info/backref.html" rel="nofollow noreferrer">here</a>的更多信息。你知道吗</p>
<pre><code># Pattern describing regular expression
pattern = re.compile(r'(\(https?[:_%A-Z=?/a-z0-9.-]+\))')
# List where we store all URLs
urls = []
# For each invoice pattern you find in string, append it to list
for url in pattern.finditer(string):
urls.append(url.group(1))
</code></pre>
<p><strong>注意:</strong></p>
<p>您应该使用<code>pattern.finditter()</code>,因为这样您可以通过调用<code>string</code>的文本中的所有模式结果进行迭代。从<em>重新查找</em>文件:</p>
<blockquote>
<p>re.finditer(pattern, string, flags=0)
Return an iterator yielding
MatchObject instances over all non-overlapping matches for the RE
pattern in string. The string is scanned left-to-right, and matches
are returned in the order found. Empty matches are included in the
result unless they touch the beginning of another match.</p>
</blockquote>