擅长:python、mysql、java
<p>要从<code>timedetla</code>对象获取分钟数,可以使用<code>total_seconds()</code>并除以60:</p>
<pre><code>epgDurationMin = epgDuration.total_seconds()/60.
if 0.10 <= epgDurationMin <= 0.30:
...
</code></pre>
<p>还请注意,您可以使用python的cool<a href="https://docs.python.org/2/reference/expressions.html#not-in" rel="nofollow">comparison-chaining</a>(例如<code>a <= b <= c</code>)</p>