Posted in Python onAugust 06, 2014
直接上代码:
#!/usr/bin/python # Filename s5.py # Python Dynamic Socks5 Proxy # Usage: python s5.py 1080 # Background Run: nohup python s5.py 1080 & import socket, sys, select, SocketServer, struct, time class ThreadingTCPServer(SocketServer.ThreadingMixIn, SocketServer.TCPServer): pass class Socks5Server(SocketServer.StreamRequestHandler): def handle_tcp(self, sock, remote): fdset = [sock, remote] while True: r, w, e = select.select(fdset, [], []) if sock in r: if remote.send(sock.recv(4096)) <= 0: break if remote in r: if sock.send(remote.recv(4096)) <= 0: break def handle(self): try: pass # print 'from ', self.client_address nothing to do. sock = self.connection # 1. Version sock.recv(262) sock.send("\x05\x00"); # 2. Request data = self.rfile.read(4) mode = ord(data[1]) addrtype = ord(data[3]) if addrtype == 1: # IPv4 addr = socket.inet_ntoa(self.rfile.read(4)) elif addrtype == 3: # Domain name addr = self.rfile.read(ord(sock.recv(1)[0])) port = struct.unpack('>H', self.rfile.read(2)) reply = "\x05\x00\x00\x01" try: if mode == 1: # 1. Tcp connect remote = socket.socket(socket.AF_INET, socket.SOCK_STREAM) remote.connect((addr, port[0])) pass # print 'To', addr, port[0] nothing do to. else: reply = "\x05\x07\x00\x01" # Command not supported local = remote.getsockname() reply += socket.inet_aton(local[0]) + struct.pack(">H", local[1]) except socket.error: # Connection refused reply = '\x05\x05\x00\x01\x00\x00\x00\x00\x00\x00' sock.send(reply) # 3. Transfering if reply[1] == '\x00': # Success if mode == 1: # 1. Tcp connect self.handle_tcp(sock, remote) except socket.error: pass #print 'error' nothing to do . except IndexError: pass def main(): filename = sys.argv[0]; if len(sys.argv)<2: print 'usage: ' + filename + ' port' sys.exit() socks_port = int(sys.argv[1]); server = ThreadingTCPServer(('', socks_port), Socks5Server) print 'bind port: %d' % socks_port + ' ok!' server.serve_forever() if __name__ == '__main__': main()
Python写的Socks5协议代理服务器
- Author -
junjie声明:登载此文出于传递更多信息之目的,并不意味着赞同其观点或证实其描述。
Reply on: @reply_date@
@reply_contents@