擅长:python、mysql、java
<p>这对我有效:</p>
<pre><code>import re
thestring = "text1\ntext2\nhttp://url.com/bla1/blah1/\ntext3\ntext4\nhttp://url.com/bla2/blah2/\ntext5\ntext6"
URLless_string = re.sub(r'\w+:\/{2}[\d\w-]+(\.[\d\w-]+)*(?:(?:\/[^\s/]*))*', '', thestring)
print URLless_string
</code></pre>
<p>结果:</p>
<pre><code>text1
text2
text3
text4
text5
text6
</code></pre>