[python] Testing socket connection in Python

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.

12 years later for anyone having similar problems.

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

doesn't work because the port '80' is a string. Your port needs to be int.

try:
    s.connect((address, 80))

This should work. Not sure why even the best answer didnt see this.

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')

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')

12 years later for anyone having similar problems.

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

doesn't work because the port '80' is a string. Your port needs to be int.

try:
    s.connect((address, 80))

This should work. Not sure why even the best answer didnt see this.

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')

You can use the function connect_ex. It doesn't throw an exception. Instead of that, returns a C style integer value (referred to as errno in C):

s =  socket.socket(socket.AF_INET, socket.SOCK_STREAM)
result = s.connect_ex((host, port))
s.close()
if result:
    print "problem with socket!"
else:
    print "everything it's ok!"