擅长:python、mysql、java
<p>这样做可以:</p>
<pre><code>s = """Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nam et nibh augue. Sed dignissim eu odio nec efficitur. Nulla aliquam imperdiet ipsum, eu mollis lacus cursus quis. Nulla dictum sem sem in auctor erat imperdiet sed suscipit elit ut lacus vestibulum vitae consequat risus volutpat. Suspendisse suscipit velit id."""
result = "\n".join(re.findall(r"(.{,59}?(?:,|\.)|.{58}\S*)\s*", s))
print(result)
</code></pre>
<p>结果:</p>
<pre><code>Lorem ipsum dolor sit amet,
consectetur adipiscing elit.
Nam et nibh augue.
Sed dignissim eu odio nec efficitur.
Nulla aliquam imperdiet ipsum,
eu mollis lacus cursus quis.
Nulla dictum sem sem in auctor erat imperdiet sed suscipit
elit ut lacus vestibulum vitae consequat risus volutpat.
Suspendisse suscipit velit id.
</code></pre>
<p><a href="https://docs.python.org/3.5/library/re.html" rel="nofollow">^{<cd1>}</a>说明:</p>
<p><code>.{,59}?(?:,|\.)</code>匹配前面少于59个字符的任何<code>,</code>或<code>.</code>。<code>.{58}\S*</code>匹配任何超过58个字符的字符,直到下一个单词。
最后,<code>\s*</code>匹配任何空空间,以便将其修剪掉。你知道吗</p>