(Guessing Game) Can't Break a loop on Python

I was trying to make a guessing game that keep playing if you choose that, but I'm having a problem to break the loop, every time the code reach "quit()", the program give me an error

when I used repl:
repl process died unexpetily

when I used programiz.com:
Traceback (most recent call last): File "<string>", line 42, in <module> File "<string>", line 40, in game File "<frozen _sitebuiltins>", line 26, in __call__ SystemExit: None

so I need some hand here, here is the code

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import random
import time

Name = input("Hello there!, What is your name? ")

print(f"Hi {Name}! Welcome to the guessing game. I'm going to pick a number between 1 and 100")

def game():
  time.sleep(2)
  print('picking ...')
  time.sleep(2)

  guess = int(input("Alright! What is your guess?: "))
  correct_number = random.randint(1,2)
  guess_count = 1

  while guess != correct_number:
    guess_count += 1
    if guess < correct_number:
      guess = int(input("Nope.Try higher. What is your guess?: "))
    else:
      guess = int(input("Nope.Try lower. What is your guess?: "))

  print(f"Congrats {Name}! The right answer was {correct_number}. It took you {guess_count} guess.")

  time.sleep(3)

  Next_Try = input('''
    Wanna Try again?
  1: Yes!
  2: No thanks
  ''')
  
  while Next_Try:
    if Next_Try == "1":
      game()
    else:
     time.sleep(2)
     print("Thanks for Playing!")
     quit()

game()

I have tried exit() but still having the same problem

In Python, exit and quit basically are both ways to immediately kill the program. In general, they should be used only when things go wrong. In your code, on the Next_Try loop at the end, you should use the keyword break. That is the proper way to exit a loop. Try this:

  while Next_Try:
    if Next_Try == "1":
      game()
    else:
     time.sleep(2)
     print("Thanks for Playing!")
     break

Here's a link to the Python docs that go over more about the break keyword.

have you tried using break instead of exit()?

while Next_Try:
    if Next_Try == "1":
      game()
    else:
     time.sleep(2)
     print("Thanks for Playing!")
     break

The code you provided appears to be a guessing game where the user tries to guess a randomly generated number between 1 and 2. While the code seems to work correctly, there are a few improvements that can be made:

The variable correct_number should generate a random number between 1 and 100 instead of 1 and 2. Update the line to correct_number = random.randint(1, 100).

Instead of using the exit() function to terminate the program, it's better to wrap the game logic in a loop and provide an option for the user to exit. This way, the program will continue running until the user explicitly chooses to exit.

Here's an updated version of your code with the mentioned improvements:

import random
import time
def game():
  Name = input("Hello there! What is your name? ")
  print(f"Hi {Name}! Welcome to the guessing game. I'm going to pick a number between 1 and 100")
  while True:
    time.sleep(2)
    print('Picking a number...')
    time.sleep(2)
    correct_number = random.randint(1, 100)
    guess_count = 0
    while True:
      guess = int(input("Alright! What is your guess?: "))
      guess_count += 1
      if guess < correct_number:
        print("Nope. Try higher.")
      elif guess > correct_number:
        print("Nope. Try lower.")
      else:
        print(f"Congrats {Name}! The right answer was {correct_number}. It took you {guess_count} guess.")
        break
    time.sleep(3)
    next_try = input('''
      Wanna try again?
      1: Yes!
      2: No thanks
    ''')
    if next_try != "1":
      break
  print("Thanks for playing!")
game()

In this updated code, the game logic is enclosed within a while True loop. It keeps the game running until the user chooses to exit by entering something other than "1" when prompted to try again.

I hope this helps! Let me know if you have any further questions or concerns.