Testing socket connection in Python
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.
Answers:
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.
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')
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!"
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.
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.
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.
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')
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!"
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.