Testing socket connection in Python

Question

This question will expand on: http://stackoverflow.com/questions/68774/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()
port = 80 # port number is a number, not string
try:
    s.connect((address, port)) 
except Exception, e:
    alert('something\'s wrong with %s:%d. Exception type is %s' % (address, port, `e`))

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 diffrently for diffrent kind of problems.

Answer 2

You should really post:

1. The complete source code of your example
2. The actual result of it, not a summary

Here is my code, which works:

import socket, sys

def alert(msg):
    print >>sys.stderr, msg
    sys.exit(1)

(family, socktype, proto, garbage, address) = \
         socket.getaddrinfo("::1", "http")[0] # Use only the first tuple
s = socket.socket(family, socktype, proto)

try:
    s.connect(address) 
except Exception, e:
    alert("Something's wrong with %s. Exception type is %s" % (address, e))

When the server listens, I get nothing (this is normal), when it doesn't, I get the expected message:

Something's wrong with ('::1', 80, 0, 0). Exception type is (111, 'Connection refused')