server_with_filter_address
3 years ago in Python
'''
TCP server and HTTP server
HHHHH
it's a test
the next project is a command line like metasploit
this project had may errors
'''
import socket
import ftplib
import string
import signal
import struct
import select
import urllib
import io, re
import os, sys
import telnetlib, ssl
import http.server, http.client
import time, datetime, uuid, uu
import selectors
import __future__
socket.hostflags=0xffff
socket.server_address=('192.168.1.5', 153)
socket.log_path='home\\var\\logs\\logs.log'
socket.filter_path='home\\var\\block\\address.ip'
socket.main_path='home\\main.html'
mode="a+"
def stdout(stdout=""):return sys.stdout.write(stdout)
def stderr(stderr=""):return sys.stderr.write(stderr)
class stdin:
def read(stdin=1):return sys.stdin.read(stdin)
def readline(stdin=1):return sys.stdin.readline(stdin)
def readlines(stdin=1):return sys.stin.readlines(stdin)
def exit(code):return sys.exit(code)
def debug_print(text="", file=sys.stdout, end="\n"):return print(text, file=file, end=end)
#class socket:pass
def start(message):
debug_print("start on %s %s" % (socket.server_address[0], socket.server_address[1]))
socket.server_socket=socket.socket(socket.AF_INET, socket.SOCK_STREAM, 0)
socket.server_socket.bind(socket.server_address)
socket.server_socket.listen(socket.hostflags)
socket.client_socket, socket.client_address=socket.server_socket.accept()
try:
socket.client_useragent=socket.server_socket.recv(socket.hostflags)
except:
socket.server_socket.send(message.encode())
debug_print("connection sucsess on: "+socket.client_address[0]+':'+str(socket.client_address[1]))
pass
with open(socket.filter_path, mode) as filter_mode:
if socket.client_address[0] in filter_mode:
socket.client_socket.recv(socket.hostflags)
socket.client_socket.send(message.encode())
else:
socket.client_socket.send(message.encode())
socket.client_socket.close()
socket.server_socket.close()
debug_print("connection sucsess on: "+socket.client_address[0]+':'+str(socket.client_address[1]))