while loop printing everything twice, Python --checking if an input is an intiger AND in range (0-10)--

You are evaluating the responses in the wrong order to be handled by the while statement. Here is one way:

def user_choice():
    choice_is_digit = False
    within_range = False

    while choice_is_digit == False or within_range == False:

        choice = input("Please enter a number (0-10): ")

        choice_is_digit =  choice.isdigit()

        if choice_is_digit == False:
            print("Please enter a digit!")
            continue  # skip rest of loop

        within_range = int(choice) in range(0,10)

        if within_range == False:
            print("Please enter a number in range (0-10)")

    return int(choice)

Having multiple if statements means that it's possible the code will hit all 3 depending on the conditions.

Changing them to an if else block means it can only use one of them.

It's checks for a condition match from the top down so if it finds the conditions match in the first if then It would skip the remaining 2 options.

Your way, it would check the first if, and if it evaluates to True it would fire the code in the if block and then check the second if statement etc etc and so on.

Try this:

`def user_choice():

choice = "wrong"
within_range = False

while choice.isdigit() == False or within_range == False:

    choice = input("Please enter a number (0-10): ")

    if choice.isdigit() == False:
        print("Please enter a digit!")

    elif within_range == False:
        print("Please enter a number in range (0-10)")

    elif choice.isdigit() == True:
        within_range = int(choice) in range(0,10)



return int(choice)`