Question

This question will expand on: Best way to open a socket in Python
When opening a socket how can I test to see if it has been established, and that it did not timeout, or generally fail.

Edit: I tried this:

try:     s.connect((address, '80')) except:     alert('failed' + address, 'down') 

but the alert function is called even when that connection should have worked.

Answer 1

It seems that you catch not the exception you wanna catch out there 🙂

if the s is a socket.socket() object, then the right way to call .connect would be:

import socket s = socket.socket() address = '127.0.0.1' port = 80  # port number is a number, not string try:     s.connect((address, port))      # originally, it was      # except Exception, e:      # but this syntax is not supported anymore.  except Exception as e:      print('something's wrong with %s:%d. Exception is %s' % (address, port, e)) finally:     s.close() 

Always try to see what kind of exception is what you’re catching in a try-except loop.

You can check what types of exceptions in a socket module represent what kind of errors (timeout, unable to resolve address, etc) and make separate except statement for each one of them – this way you’ll be able to react differently for different kind of problems.