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A request to send or receive data was disallowed because the socket is not connected in Python

I'm trying to make a console chat app in python using socket library.
Whenever I send a message to the server, the server code crashes with the following message:
OSError: [WinError 10057] A request to send or receive data was disallowed because the socket is not connected and (when sending on a datagram socket using a sendto call) no address was supplied
Server code
import socket
HOST = socket.gethostbyname(socket.gethostname()) # get the ip address of PC
PORT = 5050
ADDRESS = (HOST, PORT)
socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
socket.bind(ADDRESS)
while True:
socket.listen()
conn, addr = socket.accept()
print(f"Connected by {addr}")
while True:
data = conn.recv(64)
print(data.decode('utf-8'))
socket.send(data)
Client code
import socket
HOST = socket.gethostbyname(socket.gethostname()) # get the ip address of PC
PORT = 5050
ADDRESS = (HOST, PORT)
socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
socket.connect(ADDRESS)
while True:
msg = input("Enter your message")
socket.send(msg.encode('utf-8'))
data = socket.recv(64)
print(data.decode('utf-8'))
What I am trying to achieve is whenever I send a message to the server, the client script should print the sent message. How can I fix that?

You're attempting to send data to your own server socket. You instead want to send to the client that you accepted.
socket.send(data)
Should be:
conn.send(data)
If you think about it, if you had multiple clients, how would you send data to a specific client? By using the socket that accept gave you.
As a side note, you probably don't want to import the module as socket, and also call your variable socket. It's fine here, but if you were to make a more complicated project, you may accidentally refer to the object when you meant to refer to the module. I'd rename the socket object to sock or server_socket to avoid shadowing.

How to send and receive message from client to client with Python (socket)?

We are working on a project "ByZantine Generals Problem" with Python(socket), we manage to create a successful connection between the server and the two clients (client1, client2). But we didn't know how to create a connection between the two clients , any help ?
Link model project problem : https://upload.wikimedia.org/wikipedia/commons/thumb/8/81/4generalestenientetraidor.svg/400px-4generalestenientetraidor.svg.png
Server.py
import socket
host = '192.168.43.209' # Standard loopback interface address
(localhost)
port = 65432 # Port to listen on (non-privileged ports are > 1023)
serv = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
serv.bind((host, port))
serv.listen(5)
while True:
conn, addr = serv.accept()
conn.send(b"Attack ")
data = conn.recv(4096)
if not data: break
print (data)
client1.py
import socket
host = '192.168.43.209' # Standard loopback interface address
(localhost)
port = 65432 # Port to listen on (non-privileged ports are > 1023)
client = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
client.connect((host, port))
from_server = client.recv(4096)
print (from_server)
client.send(b"I am client 1 : ")
client2.py
import socket
host = '192.168.43.209' # Standard loopback interface address
(localhost)
port = 65432 # Port to listen on (non-privileged ports are > 1023)
client = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
client.connect((host, port))
from_server = client.recv(4096)
print (from_server)
client.send(b"I am client 2 : ")

There are two approaches to make client 1 and client 2 communicate together.
Make them communicate through the server
This means they will communicate by just connecting to the server. And the server will forward the message between them.
To make this work, you need to make a function and pass their socket object to the function when they connect. In the function, you will just receive data from the client. Every time there is data received, you will broadcast it to all other clients.
Tip: You can add each client's socket object to a list so that you can easily broadcast the message to each client in the network.
Peer to peer communication
To communicate in p2p, they don't need to connect to the server. They will just communicate with each other directly. The preferred protocol for p2p communication is, UDP protocol.
If clients are gone exchange secure data like the server shouldn't access it, p2p is the best approach. Because there is no interference of the server while they are communicating.

You can do client to client communication through the server with something like this. Note: This is not currently tested because I am not on a computer where I can run this:
The core of this code is from this answer which explains how to send a message to ALL clients: https://stackoverflow.com/a/27139338/8150685
I used a list for clients but you may find it easier to use a dictionary.
clients = [] # The clients we have connected to
clients_lock = threading.Lock()
def listener(client, address):
print "Accepted connection from: ", address
with clients_lock:
clients.append(client) # Add a client to our list
try:
while True:
data = client.recv(1024)
if not data:
break
else:
print repr(data)
# Here you need to read your data
# and figure out who you want to send it to
client_to_send_to = 1 # Send this data to client 1
with clients_lock:
if client_to_send_to < len(clients):
clients[client_to_send_to].sendall(data)
finally:
with clients_lock:
clients.remove(client)
client.close()

How to make 2 clients connect each other directly, after having both connected a meeting-point server?

I'm writing a toy meeting-point/relay server listening on port 5555 for two clients "A" and "B".
It works like this: every byte received by the server from the firstly-connected client A will be sent to the secondly-connected client B, even if A and B don't know their respective IP:
A -----------> server <----------- B # they both connect the server first
A --"hello"--> server # A sends a message to server
server --"hello"--> B # the server sends the message to B
This code is currently working:
# server.py
import socket, time
from threading import Thread
socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
socket.bind(('', 5555))
socket.listen(5)
buf = ''
i = 0
def handler(client, i):
global buf
print 'Hello!', client, i
if i == 0: # client A, who sends data to server
while True:
req = client.recv(1000)
buf = str(req).strip() # removes end of line
print 'Received from Client A: %s' % buf
elif i == 1: # client B, who receives data sent to server by client A
while True:
if buf != '':
client.send(buf)
buf = ''
time.sleep(0.1)
while True: # very simple concurrency: accept new clients and create a Thread for each one
client, address = socket.accept()
print "{} connected".format(address)
Thread(target=handler, args=(client, i)).start()
i += 1
and you can test it by launching it on a server, and do two netcat connections to it: nc <SERVER_IP> 5555.
How can I then pass the information to the clients A and B that they can talk directly to each other without making the bytes transit via the server?
There are 2 cases:
General case, i.e. even if A and B are not in the same local network
Particular case where these two clients are in the same local network (example: using the same home router), this will be displayed on the server when the 2 clients will connect to the server on port 5555:
('203.0.113.0', 50340) connected # client A, router translated port to 50340
('203.0.113.0', 52750) connected # same public IP, client B, router translated port to 52750
Remark: a previous unsuccesful attempt here: UDP or TCP hole punching to connect two peers (each one behind a router)
and UDP hole punching with a third party

Since the server knows the addresses of both clients, it can send that information to them and so they would know each others adress. There are many ways the server can send this data - pickled, json-encoded, raw bytes. I think the best option is to convert the address to bytes, because the client will know exactly how many bytes to read: 4 for the IP (integer) and 2 for the port (unsigned short). We can convert an address to bytes and back with the functions below.
import socket
import struct
def addr_to_bytes(addr):
return socket.inet_aton(addr[0]) + struct.pack('H', addr[1])
def bytes_to_addr(addr):
return (socket.inet_ntoa(addr[:4]), struct.unpack('H', addr[4:])[0])
When the clients receive and decode the address, they no longer need the server, and they can establish a new connection between them.
Now we have two main otions, as far as I know.
One client acts as a server. This client would close the connection to the server and would start listening on the same port. The problem with this method is that it will only work if both clients are on the same local network, or if that port is open for incoming connections.
Hole punching. Both clients start sending and accepting data from each other simultaneously. The clients must accept data on the same address they used to connect to the rendezvous server, which is knwn to each other. That would punch a hole in the client's nat and the clients would be able to communicate directly even if they are on different networks. This proccess is expleined in detail in this article Peer-to-Peer Communication Across Network Address Translators, section 3.4 Peers Behind Different NATs.
A Python example for UDP Hole Punching:
Server:
import socket
def udp_server(addr):
soc = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
soc.bind(addr)
_, client_a = soc.recvfrom(0)
_, client_b = soc.recvfrom(0)
soc.sendto(addr_to_bytes(client_b), client_a)
soc.sendto(addr_to_bytes(client_a), client_b)
addr = ('0.0.0.0', 4000)
udp_server(addr)
Client:
import socket
from threading import Thread
def udp_client(server):
soc = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
soc.sendto(b'', server)
data, _ = soc.recvfrom(6)
peer = bytes_to_addr(data)
print('peer:', *peer)
Thread(target=soc.sendto, args=(b'hello', peer)).start()
data, addr = soc.recvfrom(1024)
print('{}:{} says {}'.format(*addr, data))
server_addr = ('server_ip', 4000) # the server's public address
udp_client(server_addr)
This code requires for the rendezvous server to have a port open (4000 in this case), and be accessible by both clients. The clients can be on the same or on different local networks. The code was tested on Windows and it works well, either with a local or a public IP.
I have experimented with TCP hole punching but I had limited success (sometimes it seems that it works, sometimes it doesn't). I can include the code if someone wants to experiment. The concept is more or less the same, both clients start sending and receiving simultaneously, and it is described in detail in Peer-to-Peer Communication Across Network Address Translators, section 4, TCP Hole Punching.
If both clients are on the same network, it will be much easier to communicate with each other. They would have to choose somehow which one will be a server, then they can create a normal server-client connection. The only problem here is that the clients must detect if they are on the same network. Again, the server can help with this problem, as it knows the public address of both clients. For example:
def tcp_server(addr):
soc = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
soc.bind(addr)
soc.listen()
client_a, addr_a = soc.accept()
client_b, addr_b = soc.accept()
client_a.send(addr_to_bytes(addr_b) + addr_to_bytes(addr_a))
client_b.send(addr_to_bytes(addr_a) + addr_to_bytes(addr_b))
def tcp_client(server):
soc = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
soc.connect(server)
data = soc.recv(12)
peer_addr = bytes_to_addr(data[:6])
my_addr = bytes_to_addr(data[6:])
if my_addr[0] == peer_addr[0]:
local_addr = (soc.getsockname()[0], peer_addr[1])
... connect to local address ...
Here the server sends two addresses to each client, the peer's public address and the client's own public address. The clients compare the two IPs, if they match then they must be on the same local network.

The accepted answer gives the solution. Here is some additional information in the case "Client A and Client B are in the same local network".
This situation can indeed be detected by the server if it notices that both clients have the same public IP.
Then the server can choose Client A as "local server", and Client B as "local client".
The server will then ask Client A for its "local network IP". Client A can find it with:
import socket
localip = socket.gethostbyname(socket.gethostname()) # example: 192.168.1.21
and then send it back to the server. The server will communicate this "local network IP" to Client B.
Then Client A will then run a "local server":
import socket
soc = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
soc.bind(('0.0.0.0', 4000))
data, client = soc.recvfrom(1024)
print("Connected client:", client)
print("Received message:", data)
soc.sendto(b"I am the server", client)
and Client B will run as a "local client":
import socket
soc = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
server = ('192.168.1.21', 4000) # this "local network IP" has been sent Client A => server => Client B
soc.sendto("I am the client", server)
data, client = soc.recvfrom(1024)
print("Received message:", data)

Python: Socket.error Connection Refused in Unix [Errno 111]

I am trying to send UDP video packets using sockets in Python.
The Server IP address is :192.168.67.14
The Client IP address is 192.168.67.42
The Client and Server can ping each other. Below is the code used for establishing the socket:
Server Side:
import urllib, time, os, m3u8
from socket import *
# Socket initialization
s = socket(AF_INET, SOCK_DGRAM)
host = "192.168.67.42"
port = 5000
buf = 1024
addr = (host, port)
s.connect((host, port))
ts_filenames = []
while True:
playlist = "https://sevenwestmedia01-i.akamaihd.net/hls/live/224853/TEST1/master_lowl.m3u8"
m3u8_obj = m3u8.load(playlist)
ts_segments = m3u8_obj.__dict__.values()[6]
ts_segments_str = str(m3u8_obj.segments)
for line in ts_segments_str.splitlines():
if "https://" in line:
ts_id = line[-20:]
if ts_id not in ts_filenames:
print ts_id
ts_filenames.append(ts_id)
try:
ts_segment = urllib.URLopener()
ts_segment.retrieve(line, ts_id)
except:
pass
f = open(ts_id, "rb")
data = f.read(buf)
while (data):
if (s.sendto(data, addr)):
print "sending ..."
data = f.read(buf)
Client Side
import socket
s = socket.socket()
host = '192.168.67.14'
port = 5000
s.connect((host,port))
print s.recv(1024)
s.close
Exception I get:
Traceback (most recent call last): File "client.py", line 7, in
s.connect((host,port)) File "/usr/lib/python2.7/socket.py", line 228, in meth
return getattr(self._sock,name)(*args) socket.error: [Errno 111] Connection refused
I spent some time looking into this discussion but I still not sure what to modify. Any suggestions please ?

You have multiple problems here. First, by using connect on the server end, you're telling the operating system that you will only be communicating with IP address "192.168.67.42" port 5000. That is probably not what you intended. (A server usually talks to whatever client wants to talk to it.)
Second, by not specifying SOCK_DGRAM in your client, you're getting the default socket type, which is SOCK_STREAM. That means your client is trying to connect to your server on TCP port 80 -- not UDP port 80 (the two namespaces are totally separate).
For a UDP "session", both sides need an IP address and a port number. If you do not bind a port specifically, the operating system will choose one for you quasi-randomly. In order to link up client and server, they must agree on at least one of those.
So a typical UDP server will bind to a well-known port (presumably you intended 5000 for that purpose). Then the client can connect to the server at that port. The code would look something like this (sans error handling):
Server side:
# Create socket
s = socket(AF_INET, SOCK_DGRAM)
# Bind to our well known port (leave address unspecified
# allowing us to receive on any local network address)
s.bind(('', 5000))
# Receive from client (so we know the client's address/port)
buffer, client_addr = s.recvfrom(1024)
# Now we can send to the client
s.sendto(some_buffer, client_addr)
The client is close to what you have, but you should send some data from the client to the server first so that the server knows your address:
s = socket(AF_INET, SOCK_DGRAM)
# Create connection to server (the OS will automatically
# bind a port for the client)
s.connect((host, port))
# Send dummy data to server so it knows our address/port
s.send(b'foo')
buffer = s.recv(1024)
Note that because you have used connect on the client side, you've permanently specified your peer's address and don't need to use recvfrom and sendto.

On the client side, this is wrong:
s = socket.socket()
for receiving UDP packets, you need to create a UDP socket, same as you did on the server side:
s = socket(AF_INET, SOCK_DGRAM)
Also, if you want your client to be able to receive UDP packets you will need to bind() it to port 5000 (connect() is neither necessary nor sufficient for that).

Python connected datagram socket

I'm building a simple client-server multiplayer game and I want to have connected UDP sockets. However, when I call the listen() method it produces Operation not supported exception.
try:
server = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
server.bind((host, port))
server.listen(5)
except socket.error, (value, message):
print "E: " + message # E: Operation not supported
Is there a way to have connected datagram sockets?

UDP protocol is connectionless and thus you really cannot create connections between 2 sockets in the same manner as with the TCP client and server; thus you also cannot do the listen system call on an UDP socket, as it concerns only the TCP server sockets. Instead you use socket.recvfrom to receive datagrams from any address:
sock = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
sock.bind((host, port))
data, addr = sock.recvfrom(65536) # this will fit the maximum datagram
You can respond to the client (if they still have a socket open), by destructuring the addr which is a host, port tuple for socket.AF_INET, socket.SOCK_DGRAM.
client_host, client_port = addr
You can send data back there with sendto
sock.sendto(data, (client_host, client_port))
If you want to use a well-known port, you will bind the socket; if not, the system will assign a port for you.
It is possible to do a connect system call with datagram sockets on many socket implementations; this serves as a sort of filter, that drops the packets coming from unwanted sources, and for setting the default outgoing address for sock.send (it is still OK to use sock.sendto on such socket to another address, though the response might be dropped because of the "connection"). This is very useable on a client or between 2 nodes if they agree to use 2 well known ports with each other.
However if you do connect on server, it cannot serve any other requests on this socket. Also, listen with its queues only relates to SOCK_STREAM sockets.
Thus in case of many clients to 1 server, you could have the server listen to socket 12345, and when a client contacts the server, the server could respond from server:12345 with a message that contains another port number that the client should use with the server.