Catch socket timeout exception

Question

I would like to catch the socket timeout (preferably in an exception) ... except urllib.error.URLError: can catch it but I need to distinguished between a dead link and a timeout .... If I take out the except urllib.error.URLError: the socket timeout does not catch and script terminates with an socket.timeout error

import urllib.request,urllib.parse,urllib.error
import socket
import http
socket.setdefaulttimeout(0.1)


try:
    file2 = urllib.request.Request('http://uk.geforce.com/html://')
    file2.add_header("User-Agent","Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US) AppleWebKit/525.13 (KHTML, like Gecko) Chrome/0.2.149.29 Safari/525.13")
    file3 = urllib.request.urlopen(file2).read().decode("utf8", 'ignore')
except urllib.error.URLError: print('fail')
except socket.error: print('fail')
except socket.timeout: print('fail')
except UnicodeEncodeError: print('fail')
except http.client.BadStatusLine: print('fail')
except http.client.IncompleteRead: print('fail')
except urllib.error.HTTPError: print('fail')

print('done')

Answer 1

...
except urllib.error.URLError, e:
    print type(e.reason)

You will see <class 'socket.timeout'> whenever there is a socket timeout. Is this what you want?

For example:

try:
  data = urllib2.urlopen("http://www.abcnonexistingurlxyz.com")
except Exception,e:
  print type(e.reason)
... 
<class 'socket.timeout'>