如何关闭urllib/urllib 2响应消息？

Question

我用python编写了一个程序来自动从网站上检索信息。但是它会返回大量的响应消息，所以我无法看到我的程序是如何处理的。我怎么才能关掉它？

部分代码

cookiejar = cookielib.LWPCookieJar()
cookie_support = urllib2.HTTPCookieProcessor(cookiejar)
opener = urllib2.build_opener(cookie_support, urllib2.HTTPHandler(debuglevel=1))
urllib2.install_opener(opener)

....

url = 'http://weibo.com/2344328334/fans'
header = {'User-Agent':'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/27.0.1453.116 Safari/537.36'}
req = urllib2.Request(url = url, headers = header)
result = urllib2.urlopen(req)
text = result.read()

部分输出

send: 'GET /2344328334/fans HTTP/1.1\r\nAccept-Encoding: ...'
reply: 'HTTP/1.1 302 Moved Temporarily\r\n'
header: Server: nginx
header: Date: Tue, 10 Sep 2013 03:54:02 GMT
header: Content-Type: text/html; charset=utf-8
header: Transfer-Encoding: chunked
header: Connection: close
header: Cache-Control: no-cache, must-revalidate
header: Expires: Sat, 26 Jul 1997 05:00:00 GMT
... other messages ...

Answer 1

打开调试选项，只需删除debuglevel=1

并且我推荐anather：请求，它很容易使用，powerful.Just安装与pip install requests，支持python2和python3

是特征支持：

International Domains and URLs
Keep-Alive & Connection Pooling
Sessions with Cookie Persistence
...