Connection refused with python sockets - python

So I am trying to make a server program that will call the client program. The server client work fine if I call them myself from the command line but the connection is refused when the server calls it. Why is this not working?
This is the server code:
import socket,os
s = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
try:
os.remove("/tmp/SocketTest")
except OSError:
pass
s.bind("/tmp/SocketTest")
os.system("python compute.py")#compute is the client
#execfile('compute.py')
s.listen(1)
conn, addr = s.accept()
while 1:
data = conn.recv(1024)
if not data: break
conn.send(data)
conn.close()
This is the client code:
import socket
s = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
s.connect("/tmp/SocketTest")
s.send('Hello, world \n')
s.send('its a mighty fine day')
data = s.recv(1024)
s.close()
print 'Received', repr(data)

os.system will run the command you give it to completion, and you’re doing this before you call listen. As such, the client will try to connect to the server before it’s listening. Only once the client exits will the server move on past that line of code to actually start listening and accepting connections.
What you probably want to do is after the call to listen, but before the call to accept (which is when you start blocking), use subprocess.Popen to spawn a subprocess and do not wait on it.

I think the error is that you're calling compute.py before calling listen.
os.system will block your server until the call to python compute.py is completed.
Try subprocess.Popen to spawn the call to compute.py in parallel to your server in a non blocking manner. Callingsubprocess.Popen will launch python compute.py in a new process, and will continue executing the next line conn, addr = s.accept() )
#!/usr/bin/env python
import socket
import os
import subprocess
s = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
try:
os.remove("/tmp/SocketTest")
except OSError:
pass
s.bind("/tmp/SocketTest")
s.listen(1)
sp = subprocess.Popen(["/usr/bin/env", "python", "compute.py"])
conn, addr = s.accept()
while 1:
data = conn.recv(1024)
if not data:
break
conn.send(data)
conn.close()
That outputs:
Received 'Hello, world \nits a mighty fine day'

Related

How do I make my socket keep trying to connect even after the connection is closed?

So I am writing this program that will allow me to run commands from a different computer on the same network (my own version of ssh) in Python. I want the client program to run in the background of the target which I've already figured out the logistics to. What I would like to do is start the client program and never have to start it again but after I close the server program on the host computer, I get tons of errors. What I would like to do is after I close the host program, the client will continue to try to connect to the server program until I run it again. The code for my client program is here:
import socket
import os
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
def connect():
try:
s.connect(('localhost', 1234))
except:
connect()
while True:
connect()
while True:
try:
msg = s.recv(1024)
os.system(msg.decode("utf-8"))
except:
s.connect('localhost', 1234)
The code for my host program is here:
import socket
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind(('localhost', 1234))
s.listen(5)
def main():
while True:
clientsocket, address = s.accept()
while address != None:
message = input("Enter Message: ")
messageb = message.encode("utf-8")
clientsocket.send(messageb)
main()
Note: I will change the address from localhost when I put this on a different computer.
Never do the following to start over a function without understanding the consequences of recursion. A function calling itself is called recursion and uses stack space. This will crash if the stack hits the recursion limit.
def connect():
try:
s.connect(('localhost', 1234))
except:
connect()
Here's a solution.
client.py:
import socket
import time
while True:
s = socket.socket()
try:
print('Trying to connect...')
s.connect(('localhost',8000))
print('Connected.')
try:
while True:
data = s.recv(1024)
if not data: break # server closed connection if nothing received.
print(data)
finally:
s.close()
print('Disconnected.')
except ConnectionError: # Any type of connection error, e.g. refused, aborted, reset.
time.sleep(1)
server.py:
import socket
import time
s = socket.socket()
s.bind(('',8000))
s.listen()
while True:
c,a = s.accept()
print(f'Connected: {a}')
try:
while True:
c.send(b'message')
time.sleep(1)
except ConnectionError:
c.close()
print(f'Disconnected: {a}')
Note also that TCP is a streaming protocol. It has no concept of messages. Take the time.sleep(1) out and the messages will all run together. A proper implementation will have a protocol to extract complete messages from the byte stream such as fixed sized messages, size transmitted followed by message, delimiter bytes like newlines between messages, etc.

Python socket.accept() blocking code before call?

I am trying to learn Python sockets and have hit a snare with the socket.accept() method. As I understand the method, once I call accept, the thread will sit and wait for an incoming connection (blocking all following code). However, in the code below, which I got from https://docs.python.org/2/library/socket.html and am using localhost. I added a print('hello') to the first line of the server. Yet the print doesn't appear until after I disconnect the client. Why is this? Why does accept seem to run before my print yet after I bind the socket?
# Echo server program
import socket
print('hello') # This doesn't print until I disconnect the client
HOST = 'localhost'
PORT = 50007
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind((HOST, PORT))
s.listen(1)
conn, addr = s.accept()
print 'Connected by', addr
while 1:
data = conn.recv(1024)
if not data: break
conn.sendall(data)
conn.close()
# Echo client program
import socket
HOST = 'localhost' # The remote host
PORT = 50007 # The same port as used by the server
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect((HOST, PORT))
s.sendall('Hello, world')
data = s.recv(1024)
s.close()
print 'Received', repr(data)
You are likely using an output device on a system that Python's IO doesn't recognize as interactive. As a workaround, you can add sys.stdout.flush() after the print.
The standard output is a buffered stream, meaning that when you print something, the output sticks around in an internal buffer until you print enough data to fill the whole buffer (unlikely in a small program, the buffet is several kilobytes in size), or until the program exits, when all such buffers are automatically flushed. Normally when the output is a terminal service, the IO layer automatically switches to line buffering, where the buffer is also flushed whenever a newline character is printed (and which the print statement inserts automatically).
For some reason, that doesn't work on your system, and you have to flush explicitly. Another option is to run python -u, which should force unbuffered standard streams.

There is no timeout exception in socket

I am very confused about sockets... I have two scripts, one is server.py, and second is client.py:
server.py
import socket
server = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
server.bind(('0.0.0.0', 1235))
server.listen(1)
while True:
client, address = server.accept()
try:
client.recv(1024)
except socket.Timeouterror:
print 'timeout'
client.py
import socket
client = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
client.connect(('ip', 1235))
Why server.py script does not show an error of timeout?
You need to set the timeout for the socket if you wan to have one:
...
client, address = server.accept()
client.settimeout(10)
...
Or you can use a default timeout for all sockets.
socket.Timeouterror doesn't exist, it should be socket.timeout.
Also, you probably should close the client socket, otherwise the client will not know that the connection is closed. The timeout alone doesn't do that for you.
Some of the socket calls are blocking, by default. If nothing happens, they would block indefinitely. recv() is one of those calls. Other blocking calls are accept(), recvfrom(), read().

Python sockets - keep socket alive?

I'm having a little trouble with sockets in Python. Whenever someone connects it works fine but if they disconnect the server program closes. I want the server program to remain open after the client closes. I'm using a while True loop to keep the connection alive but once the client closes the connection the server closes it's connection.
Here is the client:
import socket, sys
conn = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
host = sys.argv[1]
port = int(sys.argv[2])
conn.connect((host, port))
print("Connected to host " + sys.argv[1])
td = 1
while td == 1:
msg = raw_input('MSG: ')
Here is the server:
import socket, sys
socket.setdefaulttimeout(150)
host = ''
port = 50005
socksize = 1024
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind((host, port))
print("Server started on port: %s" % port)
s.listen(1)
print("Now listening...\n")
conn, addr = s.accept()
while True:
print 'New connection from %s:%d' % (addr[0], addr[1])
data = conn.recv(socksize)
if not data:
break
elif data == 'killsrv':
conn.close()
sys.exit()
else:
print(data)
If a client closes a connection, you want it to close the socket.
It seems like there's a bit of a disconnect here that I'll try to elaborate on. When you create a socket, bind, and listen, you've established an open door for others to come and make connections to you.
Once a client connects to you, and you use the accept() call to accept the connection and get a new socket (conn), which is returned for you to interact with the client. Your original listening socket is still there and active, and you can still use it to accept more new connections.
Looking at your code, you probably want to do something like this:
while True:
print("Now listening...\n")
conn, addr = s.accept()
print 'New connection from %s:%d' % (addr[0], addr[1])
data = conn.recv(socksize)
if not data:
break
elif data == 'killsrv':
conn.close()
sys.exit()
else:
print(data)
Please note that this is just a starting point, and as others have suggested you probably want to use select() along with forking off processes or spawning threads to service each client.
Your code is only accepting a single connection - the loop only deals with the first accepted connection and terminates as soon as it lost. This is way your server exists:
data = conn.recv(socksize)
if not data:
break
What you will need to do is to accept several connections, while handling each of those in it's own loop. Note that it does not have to be a real loop for each socket, you can use a select-based approach to query which of the sockets has an event associated with it (data available, connection lost etc.) and then process only those sockets, all in the same loop.
You can also use a multi threaded / multi process approach, dealing with each client in it's own thread or process - I guess you won't run into scaling issues when playing around.
See:
http://docs.python.org/library/select.html
http://docs.python.org/library/multiprocessing.html

Python - Server and client problems

I'm trying to create a basic server and client script. The idea is that the client can connect to the server and execute commands. Kinda like SSH but very simple. Heres my server code:
import sys, os, socket
host = ''
port = 50103
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind((host, port))
print("Server started on port: ", port)
s.listen(1)
while (1):
conn, addr = s.accept()
print 'New connection from ', addr
try:
while True:
rc = conn.recv(2)
pipe = os.popen(rc)
rl = pipe.readlines()
fl = conn.makefile('w', 0)
fl.writelines(rl[:-1])
fl.close()
except IOError:
conn.close()
And here is my client:
import sys, socket
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
host = 'localhost'
port = input('Port: ')
s.connect((host, port))
while (1):
cmd = raw_input('$ ')
s.send(cmd)
file = s.makefile('r', 0)
sys.stdout.writelines(file.readlines())
file.close()
Here is my problem. I start the server and then run the client on the same machine. I enter the port and connect. Then I get the raw_input which is the '$'. If I type a command like 'ls' it just hangs on the client side. I have to exit the server for the client to receive the output of ls. By the way I am running Ubuntu Linux. Not sure if that matters.
When you makefile() on the socket and then use readlines() on it, it will continue until you reach an end of file, which in the socket case is that it closed from the other end.
Using makefile() in this case makes no sense to me, especially since you create it and close it after each command. Just use send() and recv() on both ends.
You probably also want to have some sort of actual "protocol" so the server tells the client "HERE COMES A RESPONSE" and "THIS IS THE END OF THE RESPONSE" so that the client knows. Otherwise it gets hard to know when to stop waiting for more response. :)
Update with an example that works:
server.py:
import sys, os, socket
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind(('', 50500))
print("Server started")
s.listen(1)
while True:
print "Accepting"
conn, addr = s.accept()
print 'New connection from ', addr
while True:
try:
rc = conn.recv(1024)
print "Command", rc
if not rc.strip():
continue
if rc.strip() == 'END':
print "Close"
conn.send("**END**")
conn.close()
break
else:
conn.send("This is the result of command %s\n" % rc)
except Exception:
conn.close()
sys.exit()
client.py
import sys, os, socket
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect(('localhost', 50500))
while True:
cmd = raw_input('$ ')
s.send(cmd)
result = s.recv(1024)
print result
if result == "**END**":
print "Ending"
break
Well for one thing you're only connecting on the client once and on the server you're closing the socket after every read.
You should take a look at this example.
http://ilab.cs.byu.edu/python/socket/echoserver.html
You're doing quite a few things incorrectly.

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