This is my first post in StackOverFlow and I'm sure that I can find for 100% percent an answer for it, but I'd really like to start my adventure with programming and know that this community might help me with it.

My questions are:

1. I wrote that first piece of code but I'd like to add here a function where the user writes the guess number bigger than the first one.

2. I am troubled with adding the additional function if for this situation - I do understand that it wouldn't suppose to be if, but elif.

Is anyone has any idea how to suppose to look-alike??

import random
top_range = input("Write a number bigger then 0: ")
if top_range.isdigit() or top_range[0] == "-":
    top_range = int(top_range)
    if top_range <= 0:
        print('Write a number bigger then 0')
        quit()
else:
    print('Write a number bot a word')
    quit()
    random_number = random.randint(0, top_range)
guesses = 0
while True:
    guesses  = 1
    user_guess = input("Guess the number : ")
    if user_guess.isdigit():
            user_guess = int(user_guess)
    else:
        print('Write another number.')
        continue
    if user_guess == random_number:
        print('You did it!')
        break
    else:
        if user_guess < random_number:
          print('The number is bigger then that')
        else:
          print('The number is smaller than that')
print('You did it in', guesses, "guesses!")

CodePudding user response：

Add a first if, then use elif for the other choices. I've also simplify a bit the first if

while True:
    guesses  = 1
    user_guess = input("Guess the number : ")
    if not user_guess.isdigit():
        print('Write another number.')
        continue

    user_guess = int(user_guess)

    if user_guess > top_range:
        print("That is higher than top range, try again")
    elif user_guess == random_number:
        print('You did it!')
        break
    elif user_guess < random_number:
        print('The number is bigger then that, try again')
    else:
        print('The number is smaller than that, try again')