问题
I'm making a request using urllib2
and the HTTPBasicAuthHandler
like so:
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()
I'm currently getting a httplib.BadStatusLine
exception when running this code. How could I go about debugging? Is there a way to see what the raw response is regardless of the unrecognized HTTP status code?
回答1:
Have you tried setting the debug level in your own HTTP handler? Change your code to something like this:
>>> 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
So the three lines to prepend are:
handler=urllib2.HTTPHandler(debuglevel=1)
opener = urllib2.build_opener(handler)
urllib2.install_opener(opener)
... the rest of your urllib2 code...
That will show the raw HTTP send / reply cycle on stderr.
Edit from comment
Does this work?
... same code as above this line
opener=urllib2.build_opener(authhandler, urllib2.HTTPHandler(debuglevel=1))
... rest of your code
来源:https://stackoverflow.com/questions/7697963/how-to-debug-urllib2-request-that-uses-a-basic-authentication-handler