<p>使用<code>urlparse.urlparse()</code>(“将URL解析为六个组件,返回一个6元组。这与URL的一般结构相对应:scheme://netloc/path;参数?query#fragment.”)和<code>urlparse.parse_qs()</code>(“解析作为字符串参数给出的查询字符串(数据类型为application/x-www-form-urlencoded)。数据作为字典返回。字典键是唯一的查询变量名,值是每个名称的值列表。“)</p>
<pre class="lang-python prettyprint-override"><code>>>> from urlparse import urlparse
>>> from urlparse import parse_qs
>>> urlparse('http//www.domain.com/path?a=1&b=2')
ParseResult(scheme='http', netloc='www.domain.com', path='/path', params='', query='a=1&b=2', fragment='')
>>> parse_result = urlparse('//www.domain.com/path?a=1&b=2')
>>> parse_qs(parse_result[4])
{'a': ['1'], 'b': ['2']}
</code></pre>
<p>您的示例可以如下所示:</p>
<pre class="lang-python prettyprint-override"><code>>>> for k,x in parse_qs(urlparse('blahdeblahdeblah?query=This is the query&time=8:30')[4]).items():
... print '%s=%s' % (k, x)
...
query=['This is the query']
time=['8:30']
</code></pre>
<p>有关<a href="http://docs.python.org/2/library/urlparse.html#urlparse.urlparse" rel="nofollow">urlparse</a>和<a href="http://docs.python.org/2/library/urlparse.html#urlparse.parse_qs" rel="nofollow">parse_qs</a>,请参阅Python文档</p>