问题
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
回答1:
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.
回答2:
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)
来源：https://stackoverflow.com/questions/53479668/how-to-make-2-clients-connect-each-other-directly-after-having-both-connected-a