What would be the correct way to implement a "Try again" message for a simple guessing game?

Question

I’m trying to implement a text that says "Try again" to appear when the player guesses incorrectly. This is an extremely bare bones "game" but I started coding yesterday and I’m trying to learn all the basic functions and methods. This is the code:

secret_number = 9
guess_limit = 3
guess_count = 0
while guess_count < guess_limit:
    guess = int(input("Guess:"))
    guess_count += 1
    if guess == secret_number:
        print("You won!")
        break
else:
    print("You lost!")

I tried using another "else" function and another "if" function but I couldn’t figure it out.

Asked By: sebas assaf

||

Source

Answer 1

add import random
and set secret_number = random.randint(0,9)

you don’t need to add the ‘else’ statement at the last line
it should be just "print("You lost!")"

Answered By: Radioactive Gaming

Answer 2

You can implement and ELSE statement, for continuing the loop:

secret_number = 9
guess_limit = 3
guess_count = 0
while guess_count < guess_limit:
    won = False
    guess = int(input("Guess:"))
    guess_count += 1

    if guess == secret_number:
        print("You won!")
        won = True
        break

    # If wrong, goes in here.
    else:
        # Just prints, and continues the loop
        print('Try again')

# After the loop, if not won, prints lost.
if not won:
    print("You lost!")

Answered By: GabrielBoehme

Answer 3

Welcome to Stack Exchange, sebas!

Here’s how I would implement the "try again" message:

secret_number = 9
guess_limit = 3
guess_count = 0
while guess_count < guess_limit:  # this could also just be `while True:` since the `if` will always break out
    guess = int(input("Guess:"))
    guess_count += 1
    if guess == secret_number:
        print("You won!")
        break
    else:
        if guess_count == guess_limit:
            print("You lost!")
            break
        else:
            print("Try again!")

As the comment says, the while statement could actually just be an infinite loop of while True: since the if statement logic will always eventually break out of the loop.

UPDATE: I initially said that the while/else wasn’t valid. Actually, it turns out that it is — thanks, blhsing, for pointing that out. I would probably leave it out, since to me it makes the "try again" bit of the logic easier for my brain, but it’s a nifty feature and could easily be used as well, like so:

secret_number = 9
guess_limit = 3
guess_count = 0
while guess_count < guess_limit:  # this could also just be `while True:` since the `if` will always break out
    guess = int(input("Guess:"))
    guess_count += 1
    if guess == secret_number:
        print("You won!")
        break
    else:
        if guess_count < guess_limit:
            print("Try again!")
else:
    print("You lost!")

Answered By: mister-sir

Answer 4

You can print 'Try again' after checking if the player guesses correctly. To avoid printing 'Try again' in the last guess, use an additional if statement:

secret_number = 9
guess_limit = 3
guess_count = 0
while guess_count < guess_limit:
    guess = int(input("Guess:"))
    guess_count += 1
    if guess == secret_number:
        print("You won!")
        break
    if guess_count < guess_limit: # add these two lines
        print('Try again')
else:
    print("You lost!")

Demo: https://replit.com/@blhsing/RaggedGleamingMenus

Answered By: blhsing

What would be the correct way to implement a "Try again" message for a simple guessing game?

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