擅长:python、mysql、java
<p>您不需要为此twisted,只要urllib就可以了。见<a href="http://pythonquirks.blogspot.com/2009/12/asynchronous-http-request.html" rel="nofollow noreferrer">http://pythonquirks.blogspot.com/2009/12/asynchronous-http-request.html</a></p>
<p>我在这里复制了相关的代码,但是这个链接是值得赞扬的:</p>
<pre>
import urllib2
class MyHandler(urllib2.HTTPHandler):
def http_response(self, req, response):
return response
o = urllib2.build_opener(MyHandler())
o.open('http://www.google.com/')
</pre>