A solution to break/continue a Python loop?

Question

The point is, I can’t find a solution that replaces using the console for my small application.

What happens is: My app runs in eternal loop, but after it runs the loop, it should stop to ask if I want to stop the loop or continue. So far nothing impossible. However, I want it to continue the loop if I do nothing. For example:

1. I open the program.
2. The program executes the first loop.
3. A message appears: "Press any key to stop or wait 4 seconds."

From that moment there are two possibilities.

1. I press any key and the program closes.
2. I don’t squeeze anything and the program continues.

I have already thought of some things, such as using two threads one to read the keys (if any) and another to count the 4 seconds, but I could not. So I come to ask if anyone knows a solution to this problem.

import pyautogui
import sys, time, msvcrt, random, time, os

# obs: Minha resolução de tela é 1366 x 768. Talvez será necessário adaptar as coordenadas
#      dos cliques automatizados para sua tela. 

def jogar():
    time.sleep(0.5)
    pyautogui.click(670, 330)
    time.sleep(0.5)
    cartaEscolhida = random.randint(1,5) # Escolhe uma carta aleatória.
    if cartaEscolhida == 1: 
        pyautogui.click(670, 330)
    elif cartaEscolhida == 2:
        pyautogui.click(717, 350)
    elif cartaEscolhida == 3:
        pyautogui.click(771, 351)
    elif cartaEscolhida == 4:
        pyautogui.click(815, 357)
    else:
        pyautogui.click(863, 350) 
    time.sleep(7.5)
    pyautogui.click(760, 410)

def abrirJogo():                # Dá os cliques para abrir a extensão e o jogo de cartas.
    pyautogui.click(891, 537)
    time.sleep(1)
    pyautogui.click(694, 386)

def fecharJogo():               # Dá os cliques para fechar a extensão.
    time.sleep(0.5)
    pyautogui.click(880, 531)

def mudarAba():                 # Para mudar da aba do console para a aba do navegador.
    pyautogui.keyDown('alt',)
    pyautogui.press('tab')
    pyautogui.keyUp('alt')

def pararPrograma():
    timeout = 4
    startTime = time.time()
    inp = None
    print("Aperte qualquer tecla para continuar ou aguarde 4 segundos.")
    while True:
        if msvcrt.kbhit():
            inp = msvcrt.getch()
            break
        elif time.time() - startTime > timeout:
            break
    if inp:
        sys.exit()
    else: 
       parar = False

clear = lambda: os.system('cls')  # Para limpar o console. 

# Daqui pra frente é o programa funcionando. 

print("Abra o canal na twitch e deixe em modo teatro. O programa iniciará em 10 segundos.")
time.sleep(10)
parar = False
while parar == False:
     abrirJogo()
     jogar()
     fecharJogo()
     mudarAba()
     clear()
     pararPrograma()     
     mudarAba()

I also put the source code on the website Pastebin.com and recorded a video running it.

Note: The current source code works, however I want to solve this problem because I want to use a graphical interface.

To simplify a little more the problem I did it:

from threading import Thread
import time, sys

def pergunta():
    print("Aperte qualquer tecla para parar ou aguarde 4 segundos.")
    input()

def contar():
    timeout = 4
    startTime = time.time()
    while True:
        if time.time() - startTime > timeout:
            print("Continuando")
            break

ThreadA = Thread(target=pergunta)
ThreadB = Thread(target=contar)

ThreadA.start()
ThreadB.start()

Now the problem changes. The input does not end after the 4 seconds. And I cannot call these threads again.

Answer 1

Good guys. It’s 4:31 in the morning and I solved. = D I’ll post on Google drive since they are almost identical codes.

The solution was I put the thread(deletes one) inside a while with a for. That way every time the thread finishes it will go back to while that will repeat the thread.

And the input I left as something normal. However now, Once 4 seconds pass and nothing is pressed, the program will type "continue" and continue the loop.

Obs: v3 is where the solution is.

The link