urllib error on urlopen

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  • Mike Driscoll
    #1

    urllib error on urlopen

    Hi,

    I have been using the following code for over a year in one of my
    programs:

    f = urllib2.urlopen ('https://www.companywebs ite.com/somestring')

    It worked great until the middle of the afternoon yesterday. Now I get
    the following traceback:

    Traceback (most recent call last):
    File "<pyshell#1 3>", line 1, in <module>
    response = urllib2.urlopen (req).read().st rip()
    File "c:\python25\li b\urllib2.py", line 124, in urlopen
    return _opener.open(ur l, data)
    File "c:\python25\li b\urllib2.py", line 381, in open
    response = self._open(req, data)
    File "c:\python25\li b\urllib2.py", line 399, in _open
    '_open', req)
    File "c:\python25\li b\urllib2.py", line 360, in _call_chain
    result = func(*args)
    File "c:\python25\li b\urllib2.py", line 1115, in https_open
    return self.do_open(ht tplib.HTTPSConn ection, req)
    File "c:\python25\li b\urllib2.py", line 1082, in do_open
    raise URLError(err)
    URLError: <urlopen error (1, 'error:140770FC :SSL
    routines:SSL23_ GET_SERVER_HELL O:unknown protocol')>


    I tried my Google Fu on this error, but there's not much out there. I
    tried using a proxy in Python, but that returned the same traceback.
    If I copy the URL into my browser, it resolves correctly. Does anyone
    have any advice on how to troubleshoot this error?

    I am using Python 2.5.2 on Windows XP.

    Thanks,

    Mike

    Tags: urllib error
    • Michael Palmer
      #2
      Re: urllib error on urlopen

      On Sep 24, 11:46 am, Mike Driscoll <kyoso...@gmail .comwrote:
      Hi,
      >
      I have been using the following code for over a year in one of my
      programs:
      >
      f = urllib2.urlopen ('https://www.companywebs ite.com/somestring')
      >
      It worked great until the middle of the afternoon yesterday. Now I get
      the following traceback:
      >
      Traceback (most recent call last):
      File "<pyshell#1 3>", line 1, in <module>
      response = urllib2.urlopen (req).read().st rip()
      File "c:\python25\li b\urllib2.py", line 124, in urlopen
      return _opener.open(ur l, data)
      File "c:\python25\li b\urllib2.py", line 381, in open
      response = self._open(req, data)
      File "c:\python25\li b\urllib2.py", line 399, in _open
      '_open', req)
      File "c:\python25\li b\urllib2.py", line 360, in _call_chain
      result = func(*args)
      File "c:\python25\li b\urllib2.py", line 1115, in https_open
      return self.do_open(ht tplib.HTTPSConn ection, req)
      File "c:\python25\li b\urllib2.py", line 1082, in do_open
      raise URLError(err)
      URLError: <urlopen error (1, 'error:140770FC :SSL
      routines:SSL23_ GET_SERVER_HELL O:unknown protocol')>
      >
      I tried my Google Fu on this error, but there's not much out there. I
      tried using a proxy in Python, but that returned the same traceback.
      If I copy the URL into my browser, it resolves correctly. Does anyone
      have any advice on how to troubleshoot this error?
      >
      I am using Python 2.5.2 on Windows XP.
      >
      Thanks,
      >
      Mike
      Could it just be a misconfiguratio n at the other end? Can you open
      other https urls?

        Comment

        • Steven D'Aprano
          #3
          Re: urllib error on urlopen

          On Wed, 24 Sep 2008 08:46:56 -0700, Mike Driscoll wrote:
          Hi,
          >
          I have been using the following code for over a year in one of my
          programs:
          >
          f = urllib2.urlopen ('https://www.companywebs ite.com/somestring')
          >
          It worked great until the middle of the afternoon yesterday. Now I get
          the following traceback:
          ....
          URLError: <urlopen error (1, 'error:140770FC :SSL
          routines:SSL23_ GET_SERVER_HELL O:unknown protocol')>
          Have you recently set a proxy where Python can auto-detect it? I
          understand that urllib2 doesn't work well with https proxies.

          If so, you can instruct urllib2 not to use a proxy-handler, but it's more
          work. What I do is construct an opener without a proxyhandler:


          # untested...
          no_proxy_suppor t = urllib2.ProxyHa ndler({})
          opener = urllib2.build_o pener(no_proxy_ support)
          f = opener.open('ht tps://www.companywebs ite.com/somestring')


          If that doesn't work, you may need to build a Request object from the URL
          before passing it to opener.open.



          --
          Steven

            Comment

            • Mike Driscoll
              #4
              Re: urllib error on urlopen

              On Sep 24, 7:08 pm, Michael Palmer <m_palme...@yah oo.cawrote:
              On Sep 24, 11:46 am, Mike Driscoll <kyoso...@gmail .comwrote:
              >
              >
              >
              Hi,
              >
              I have been using the following code for over a year in one of my
              programs:
              >
              f = urllib2.urlopen ('https://www.companywebs ite.com/somestring')
              >
              It worked great until the middle of the afternoon yesterday. Now I get
              the following traceback:
              >
              Traceback (most recent call last):
                File "<pyshell#1 3>", line 1, in <module>
                  response = urllib2.urlopen (req).read().st rip()
                File "c:\python25\li b\urllib2.py", line 124, in urlopen
                  return _opener.open(ur l, data)
                File "c:\python25\li b\urllib2.py", line 381, in open
                  response = self._open(req, data)
                File "c:\python25\li b\urllib2.py", line 399, in _open
                  '_open', req)
                File "c:\python25\li b\urllib2.py", line 360, in _call_chain
                  result = func(*args)
                File "c:\python25\li b\urllib2.py", line 1115, in https_open
                  return self.do_open(ht tplib.HTTPSConn ection, req)
                File "c:\python25\li b\urllib2.py", line 1082, in do_open
                  raise URLError(err)
              URLError: <urlopen error (1, 'error:140770FC :SSL
              routines:SSL23_ GET_SERVER_HELL O:unknown protocol')>
              >
              I tried my Google Fu on this error, but there's not much out there. I
              tried using a proxy in Python, but that returned the same traceback.
              If I copy the URL into my browser, it resolves correctly. Does anyone
              have any advice on how to troubleshoot this error?
              >
              I am using Python 2.5.2 on Windows XP.
              >
              Thanks,
              >
              Mike
              >
              Could it just be a misconfiguratio n at the other end? Can you open
              other https urls?
              This is really weird. Now it works this morning. I've spoken with our
              webmaster/system admin and he said he didn't change anything on his
              end. We're both befuddled. Sorry for the noise.

              Mike

                Comment

                • Mike Driscoll
                  #5
                  Re: urllib error on urlopen

                  On Sep 24, 9:36 pm, Steven D'Aprano <st...@REMOVE-THIS-
                  cybersource.com .auwrote:
                  On Wed, 24 Sep 2008 08:46:56 -0700, Mike Driscoll wrote:
                  Hi,
                  >
                  I have been using the following code for over a year in one of my
                  programs:
                  >
                  f = urllib2.urlopen ('https://www.companywebs ite.com/somestring')
                  >
                  It worked great until the middle of the afternoon yesterday. Now I get
                  the following traceback:
                  ...
                  URLError: <urlopen error (1, 'error:140770FC :SSL
                  routines:SSL23_ GET_SERVER_HELL O:unknown protocol')>
                  >
                  Have you recently set a proxy where Python can auto-detect it? I
                  understand that urllib2 doesn't work well with https proxies.
                  >
                  If so, you can instruct urllib2 not to use a proxy-handler, but it's more
                  work. What I do is construct an opener without a proxyhandler:
                  >
                  # untested...
                  no_proxy_suppor t = urllib2.ProxyHa ndler({})
                  opener = urllib2.build_o pener(no_proxy_ support)
                  f = opener.open('ht tps://www.companywebs ite.com/somestring')
                  >
                  If that doesn't work, you may need to build a Request object from the URL
                  before passing it to opener.open.
                  >
                  --
                  Steven
                  As I mentioned to Mr. Palmer, the error has mysteriously gone away
                  this morning. I'll keep your advice handy though, in case it happens
                  again.

                  Thanks,

                  Mike

                    Comment

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