Catch unknown exception

[Thomas Lindgaard]

Hello

Python is not my native language, so I'm kinda stuck right now. 

I have a small web crawler, that opens pages using urllib.urlopen (1). The
problem is that when a page doesn't exist, I would like to do this and
that. So I have done the following:

  import urllib
  import socket
  socket.setdefaulttimeout(10)

  def getPage(self):
    try:
      self.page = urllib.urlopen(self.link)
      self.body = self.page.read()
      self.page.close()
    except <this is the problem>:
      singSongAboutTimeout()

I have tried "except socket.timeout" but this is ignored. The call to
getPage is wrapped in its own try-except:

  try:
    self.page = getPage(self.link)
    self.body = self.page.read()
    self.page.close()
  except IOError, (errno, errmsg):
    # print error

This block catches the exception, but I would really like to catch it
earlier.

How do I do that? Is there a way to check which exception is being caught
- something along the lines of:

  try:
    # something throws an exception
  except:
    printHumanReadableDescriptionOfException()

And finally, I have a few problems finding what to write after except...
how can I know what the exception returns (ie. errno and errmsg are
returned by IOError)?

Regards
/Thomas

(1) I have specified a new user-agent as described in
http://python.org/doc/2.3.4/lib/module-urllib.html under _urlopener