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.

Asked by: Walter116 | Posted: 28-01-2022

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.

Answered by: Cherry246 | Posted: 01-03-2022

Answer 2

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!"

Answered by: Sawyer851 | Posted: 01-03-2022

Answer 3

You should really post:

The complete source code of your example
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')

Answered by: Catherine223 | Posted: 01-03-2022

Answer 4

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.

Answered by: Samantha573 | Posted: 01-03-2022

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