How to solve this User Input problem in countdown timer Python3.9.4

Question

I have a countdown timer that asks for user input for a "lucky number". Once that number is entered, the timer starts to countdown and prints out a simple message.

My problem is that if the user input is not a number, I get the "Traceback (most recent call last) ValueError: invalid literal for int() with base 10"

I understand what that means, but I don't know how to make it bypass this condition so that the user input can be anything (not just numbers) and not give me an error, but rather just print out a message that would just say; "continue without lucky number"..-and just continue with the rest of the script..

Any help would be highly appreciated. Thanks.

Here is a screenshot of the code

Answer 1

treat the user input as a string and then use a try-except to determine whether or not it is a number

try: 
  number = int(stringvar)
except:
  print("continue without lucky number")

you could also then generate a random number for your timer if the user did not enter a number

import time
import random

def countdown(t):
    while t:
        mins, secs = divmod(t, 60)
        timer = "{:02d}:{:02d}".format(mins, secs)
        print(timer, end='\r')
        time.sleep(1)
        t -= 1

num = input("Enter your lucky number: ")

isNum = False

try:
    i = int(num)
    isNum = True
except:
    print("Continue without lucky number")

if isNum:
    countdown(i)
else:
    rand = random.randint(1, 100)
    countdown(rand)
    print(f"timer number is: {rand}")

that can be your whole timer program.