擅长:python、mysql、java
<p>使用下面的regexp:</p>
<pre><code>s = """<a href="\n\n\n\n\n\n https://www.amazon.co.uk/Nikon-Coolpix-L340-Bridge-Camera/dp/B00THKEKEQ/ref=zg_bs_560836_2&#10;">Nikon Coolpix L340 Bridge Camera - Bl...</a>"""
re.findall('(?<=\n\n\n\n\n\n)(.*?)"', s)
</code></pre>
<p>以前的regexp是在字符串的开头寻找<code>\n...</code>匹配,而不是像示例字符串那样在字符串中间寻找<code>\n</code>匹配</p>