mardi 4 décembre 2018

alternative if statement to check if x matches random output

So I have written this code and I want to make the function check to see if the random.choice matches the input from the user however, I have used if x in numbers: which checks to see if x matches one of the numbers from the list. That's not what I want and so my question is, how can I make this possible. I know my code is really hard to see but right now that is not my concern

    import time
    import random

    tokens = 10
    age = int(input("Enter your age: "))
    legal = 18
    numbers = [1,2,3,4,5,6]
    x = int()
    answer = str()

    def aboveage():
        if (int(age)) >= (int(legal)):
            print("You are old enough to play")
            time.sleep(0.5)
        else:
            print("You can't play")
            if (int(age)) >= (int(legal)) and (int(tokens)) >= 10:
                answer = input("Would you like to bet?: ")
                if answer == "yes":
                    x = int(input("Choose the number the die will show: "))
                    if x in numbers:
                        print("The die is rolling...")
                        time.sleep(2)
                        print(random.choice(numbers))
                        if x in [1,2,3,4,5,6]:
                            print("You won!")
                        else:
                            print("You lost :(")
                    else:
                        print("You need to guess a number between 1-6")
                else:
                    print("Try again later")
            elif (int(age)) < (int(legal)):
                print("You are underage")
            else:
                print("You don't have enough tokens")

                if x in numbers:
                    print("Congratulations you won! :)")
                else:
                    print("You lost :(")

                    return

    aboveage()

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