Making a program than would ask the user for an integer then will asked if they wish to loop or quit

Question

def interact():
    while True: 
        try:
            num = int(input("Please input an integer: "))
            if (num % 2) == 0:
                print ("{0} is even".format(num))
            else:
                print("{0} is odd".format(num))
            num_two = int(input('Do you want to play again n/Y:'))
       
        except:
            if num_input == "y":
                continue
        finally:
            print("Goodbye")
        print(num_two)

In this code I am making a program that would ask the user for an integer then it will show if it is odd or even then will ask the user if to continue or quit.If the user enter the keyword for continue then it will ask for an integer again or else it will quit.

Answer 1

Use a break statment after your Goodbye:

def interact():
    while True: 
        try:
            num = int(input("Please input an integer: "))
            if (num % 2) == 0:
                print ("{0} is even".format(num))
            else:
                print("{0} is odd".format(num))
            num_two = int(input('Do you want to play again n/Y:'))
       
        except:
            if num_input == "y":
                continue
        finally:
            print("Goodbye")
            break
        print(num_two)

Answer 2

The play-again prompt expects a string, not a number, to be entered. Don't try to treat it as a number. Restrict the try statement to testing the number input: if you get a ValueError, restart the loop immediately, instead of trying to test if it is even or odd.

When you get the response to play again, use break to exit the loop if the response isn't y.

Print "Good bye" after the loop exits.

def interact():
    while True:
        try:
            num = int(input("Please input an integer: "))
        except ValueError:
            continue  # Ask for a number again

        if num % 2 == 0:
            print("{0} is even".format(num))
        else:
            print("{0} is odd".format(num))

        response = input("Play again? (n/Y) ")
        if response != "y":
            break
    print("Good bye")

Answer 3

I don't fully understand what you are after, but maybe try something like this:

def interact():
    while True: 
        try:
            num = int(input("Please input an integer: "))
            if (num % 2) == 0:
                print ("{0} is even".format(num))
            else:
                print("{0} is odd".format(num))
        except:
            continue
        if input("Would you like to play again? Y/N: ").lower() == "y":
            continue
        else:
            print("goodbye")
            break

I believe this will give the effect your are after.

Answer 4

A more compact approach:

def interact():
    want_continue = "Y"
    while want_continue == "Y":
        try:
            num = int(input("Please input an integer: "))
            print ("{0} is {1}".format(num, "even" if num % 2 == 0 else "odd"))
            want_continue = input('Do you want to play again (n/Y):').upper()
        except ValueError:
            print("Please enter a number")
    print("Goodbye")