Hi,
I have been using the following code for over a year in one of my
programs:
f = urllib2.urlopen('https://www.companywebsite.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#13>", line 1, in <module>
response = urllib2.urlopen(req).read().strip()
File "c:\python25\lib\urllib2.py", line 124, in urlopen
return _opener.open(url, data)
File "c:\python25\lib\urllib2.py", line 381, in open
response = self._open(req, data)
File "c:\python25\lib\urllib2.py", line 399, in _open
'_open', req)
File "c:\python25\lib\urllib2.py", line 360, in _call_chain
result = func(*args)
File "c:\python25\lib\urllib2.py", line 1115, in https_open
return self.do_open(httplib.HTTPSConnection, req)
File "c:\python25\lib\urllib2.py", line 1082, in do_open
raise URLError(err)
URLError: <urlopen error (1, 'error:140770FC:SSL
routines:SSL23_GET_SERVER_HELLO: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 4 5513
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.companywebsite.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#13>", line 1, in <module>
response = urllib2.urlopen(req).read().strip()
File "c:\python25\lib\urllib2.py", line 124, in urlopen
return _opener.open(url, data)
File "c:\python25\lib\urllib2.py", line 381, in open
response = self._open(req, data)
File "c:\python25\lib\urllib2.py", line 399, in _open
'_open', req)
File "c:\python25\lib\urllib2.py", line 360, in _call_chain
result = func(*args)
File "c:\python25\lib\urllib2.py", line 1115, in https_open
return self.do_open(httplib.HTTPSConnection, req)
File "c:\python25\lib\urllib2.py", line 1082, in do_open
raise URLError(err)
URLError: <urlopen error (1, 'error:140770FC:SSL
routines:SSL23_GET_SERVER_HELLO: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 misconfiguration at the other end? Can you open
other https urls?
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.companywebsite.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_HELLO: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_support = urllib2.ProxyHandler({})
opener = urllib2.build_opener(no_proxy_support)
f = opener.open('https://www.companywebsite.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
On Sep 24, 7:08*pm, Michael Palmer <m_palme...@yahoo.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.companywebsite.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#13>", line 1, in <module>
* * response = urllib2.urlopen(req).read().strip()
* File "c:\python25\lib\urllib2.py", line 124, in urlopen
* * return _opener.open(url, data)
* File "c:\python25\lib\urllib2.py", line 381, in open
* * response = self._open(req, data)
* File "c:\python25\lib\urllib2.py", line 399, in _open
* * '_open', req)
* File "c:\python25\lib\urllib2.py", line 360, in _call_chain
* * result = func(*args)
* File "c:\python25\lib\urllib2.py", line 1115, in https_open
* * return self.do_open(httplib.HTTPSConnection, req)
* File "c:\python25\lib\urllib2.py", line 1082, in do_open
* * raise URLError(err)
URLError: <urlopen error (1, 'error:140770FC:SSL
routines:SSL23_GET_SERVER_HELLO: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 misconfiguration 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
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.companywebsite.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_HELLO: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_support = urllib2.ProxyHandler({})
opener = urllib2.build_opener(no_proxy_support)
f = opener.open('https://www.companywebsite.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 This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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