如何调试使用基本认证处理程序的urllib2请求

Question

我正在使用 urllib2 和 HTTPBasicAuthHandler 来发送请求，代码是这样的：

import urllib2

theurl = 'http://someurl.com'
username = 'username'
password = 'password'

passman = urllib2.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, theurl, username, password)

authhandler = urllib2.HTTPBasicAuthHandler(passman)
opener = urllib2.build_opener(authhandler)
urllib2.install_opener(opener)

params = "foo=bar"

response = urllib2.urlopen('http://someurl.com/somescript.cgi', params)

print response.info()

但是当我运行这段代码时，出现了一个 httplib.BadStatusLine 的错误。我该怎么调试呢？有没有办法查看原始的响应内容，不管它的HTTP状态码是什么？

Answer 1

你有没有试过在自己的HTTP处理程序中设置调试级别？把你的代码改成下面这样：

>>> import urllib2
>>> handler=urllib2.HTTPHandler(debuglevel=1)
>>> opener = urllib2.build_opener(handler)
>>> urllib2.install_opener(opener)
>>> resp=urllib2.urlopen('http://www.google.com').read()
send: 'GET / HTTP/1.1
      Accept-Encoding: identity
      Host: www.google.com
      Connection: close
      User-Agent: Python-urllib/2.7'
reply: 'HTTP/1.1 200 OK'
header: Date: Sat, 08 Oct 2011 17:25:52 GMT
header: Expires: -1
header: Cache-Control: private, max-age=0
header: Content-Type: text/html; charset=ISO-8859-1
... the remainder of the send / reply other than the data itself 

所以要加上的三行代码是：

handler=urllib2.HTTPHandler(debuglevel=1)
opener = urllib2.build_opener(handler)
urllib2.install_opener(opener)
... the rest of your urllib2 code...

这样做会在错误输出中显示原始的HTTP发送和回复过程。

评论补充

这样做有效吗？

... same code as above this line
opener=urllib2.build_opener(authhandler, urllib2.HTTPHandler(debuglevel=1))
... rest of your code